NCERT Solutions · Class 11 Chemistry Equilibrium

A weak base's ionization constant comes from its \(\displaystyle [\text{OH}^-]\) at equilibrium, and that means going through pOH first — not pH directly.Codeine (call it B, since \(\displaystyle \text{C}_{18}\text{H}_{21}\text{NO}_3\) is unwieldy to keep rewriting) is a weak base. In water it partly ionizes:\[\text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^- \]Step $\displaystyle 1$ — get pOH from pH.Use \(\displaystyle \text{pH} + \text{pOH} = 14\) (this holds for water at $\displaystyle 298$ K, where \(\displaystyle K_w = 10^{-14}\)).\[\text{pOH} = 14 - \text{pH} = 14 - 9.95 = 4.05 \]A base gives a high pH, so it's easy to accidentally plug $\displaystyle 9.95$ straight into \(\displaystyle [\text{OH}^-] = 10^{-\text{pH}}\) — that would silently compute \(\displaystyle [\text{H}^+]\) instead. Convert to pOH first.Step $\displaystyle 2$ — get \(\displaystyle [\text{OH}^-]\) from pOH.\[[\text{OH}^-] = 10^{-\text{pOH}} = 10^{-4.05} = 10^{-4} \times 10^{-0.05} \ \text{mol L}^{-1} \]Since \(\displaystyle 10^{-0.05} = 0.891\),\[[\text{OH}^-] = 8.91 \times 10^{-5} \ \text{mol L}^{-1} \]Step $\displaystyle 3$ — set up the equilibrium table.Let \(\displaystyle C = 0.005\ \text{mol L}^{-1}\) be the starting concentration of codeine, and let \(\displaystyle x\) be how much of it ionizes.

	B	\(\displaystyle \text{BH}^+\)	\(\displaystyle \text{OH}^-\)
initial	\(\displaystyle C\)	$\displaystyle 0$	$\displaystyle 0$
at equilibrium	\(\displaystyle C-x\)	\(\displaystyle x\)	\(\displaystyle x\)

Every \(\displaystyle \text{OH}^-\) ion produced comes with one \(\displaystyle \text{BH}^+\), and both equal \(\displaystyle x\). We already found \(\displaystyle [\text{OH}^-]\) at equilibrium, so\[x = 8.91 \times 10^{-5}\ \text{mol L}^{-1} \]Step $\displaystyle 4$ — check whether \(\displaystyle C - x \approx C\) is safe.\[\frac{x}{C} = \frac{8.91\times 10^{-5}}{5\times 10^{-3}} = 0.0178 = 1.78\% \]That's under $\displaystyle 5$%, so approximating \(\displaystyle C - x \approx C\) in the denominator is justified — skipping this check is the step people get wrong, because it silently turns an approximation into an assumption.Step $\displaystyle 5$ — write \(\displaystyle K_b\) and substitute.The base ionization constant is\[K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]} = \frac{x^2}{C - x} \approx \frac{x^2}{C} \]\[K_b = \frac{(8.91\times 10^{-5}\ \text{mol L}^{-1})^2}{5\times 10^{-3}\ \text{mol L}^{-1}} = \frac{7.94\times 10^{-9}\ \text{mol}^2\text{L}^{-2}}{5\times 10^{-3}\ \text{mol L}^{-1}} \]\[K_b = 1.59 \times 10^{-6}\ \text{mol L}^{-1} \]Step $\displaystyle 6$ — convert to \(\displaystyle pK_b\).\[pK_b = -\log K_b = -\log(1.59\times 10^{-6}) = 6 - \log(1.59) = 6 - 0.20 = 5.80 \]\(\displaystyle pK_b\) has no units — it's a log of a number, not a physical quantity — but \(\displaystyle K_b\) itself is quoted in mol L\(\displaystyle ^{-1}\), since the ionization is a $\displaystyle 1$:$\displaystyle 1$:$\displaystyle 1$ reaction and the mol\(\displaystyle ^2\)L\(\displaystyle ^{-2}\) from the numerator divided by mol L\(\displaystyle ^{-1}\) leaves one power of concentration.Answer: \(\displaystyle K_b \approx 1.59 \times 10^{-6}\ \text{mol L}^{-1}\), \(\displaystyle pK_b \approx 5.80\)

Aniline is a weak base — set up its ionization equilibrium, use the fact that \(\displaystyle K_b \ll C\) to simplify, and go via [OH⁻] and pOH, since it's \(\displaystyle K_b\) you were given, not \(\displaystyle K_a\).In water, aniline accepts a proton from water to form the anilinium ion: \[C_6H_5NH_2 + H_2O \rightleftharpoons C_6H_5NH_3^+ + OH^- \]From Table $\displaystyle 6.7$, the base ionization constant of aniline is \[K_b = 4.27 \times 10^{-10} \]Let \(\displaystyle C = 0.001\ \text{M} = 1\times10^{-3}\ \text{M}\) be the initial concentration of aniline, and let \(\displaystyle \alpha\) be its degree of ionization — the fraction of the dissolved aniline that has actually reacted with water. At equilibrium the concentrations are: aniline \(\displaystyle = C(1-\alpha)\), anilinium ion \(\displaystyle = C\alpha\), hydroxide ion \(\displaystyle = C\alpha\).\[K_b = \frac{[C_6H_5NH_3^+][OH^-]}{[C_6H_5NH_2]} = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha} \]Here is the step people skip: you cannot drop \(\displaystyle (1-\alpha)\) just because it looks convenient — you drop it only after checking that \(\displaystyle \alpha\) will turn out to be tiny. Since \(\displaystyle K_b \sim 10^{-10}\) is far smaller than \(\displaystyle C \sim 10^{-3}\), \(\displaystyle \alpha\) will indeed be small, so \(\displaystyle 1-\alpha \approx 1\):\[K_b \approx C\alpha^2 \quad\Rightarrow\quad \alpha = \sqrt{\frac{K_b}{C}} \]Substituting the numbers: \[\alpha = \sqrt{\frac{4.27\times10^{-10}}{1\times10^{-3}}} = \sqrt{4.27\times10^{-7}} = 6.53\times10^{-4} \]Since \(\displaystyle \alpha = 6.53\times10^{-4} \ll 1\), the approximation \(\displaystyle 1-\alpha \approx 1\) was justified — no need to solve the quadratic.Now find the hydroxide concentration: \[[OH^-] = C\alpha = (1\times10^{-3}\ \text{mol/L})(6.53\times10^{-4}) = 6.53\times10^{-7}\ \text{mol/L} \]Because you were given \(\displaystyle \mathrm{K_b}\), this calculation naturally lands you on \(\displaystyle [OH^-]\), not \(\displaystyle [H^+]\) so find pOH first, then convert.\[\text{pOH} = -\log[OH^-] = -\log(6.53\times10^{-7}) = 7 - \log(6.53) = 7 - 0.815 = 6.19 \]At $\displaystyle 298$ K, \(\displaystyle K_w = 1.0\times10^{-14}\), so \(\displaystyle \text{pH} + \text{pOH} = 14\): \[\text{pH} = 14 - \text{pOH} = 14 - 6.19 = 7.82 \]Finding \(\displaystyle K_a\) of the conjugate acid. The conjugate acid of aniline is the anilinium ion, which itself can donate a proton back to water: \[C_6H_5NH_3^+ + H_2O \rightleftharpoons C_6H_5NH_2 + H_3O^+ \]For any conjugate acid–base pair, their ionization constants are tied together through the ionic product of water: \[K_a \times K_b = K_w = 1.0\times10^{-14}\ (\text{at }298\ \text{K}) \]This relation only holds for a genuine conjugate pair (aniline and anilinium ion here) — it is not a formula for combining two unrelated acids and bases. Solving for \(\displaystyle K_a\):\[K_a = \frac{K_w}{K_b} = \frac{1.0\times10^{-14}}{4.27\times10^{-10}} = 2.34\times10^{-5} \]Rounding each result to three significant figures, matching the precision of the given \(\displaystyle K_b\):Answer: pH ≈ $\displaystyle 7.82$; degree of ionization \(\displaystyle \alpha\) ≈ $\displaystyle 6.53$ × $\displaystyle 10$⁻⁴; ionization constant of the conjugate acid (anilinium ion) \(\displaystyle K_a\) ≈ $\displaystyle 2.34$ × $\displaystyle 10$⁻⁵

The degree of ionization of a weak acid is fixed by its dissociation constant \(\displaystyle \mathrm{K_a^{-}}\) and adding a strong acid dumps extra \(\displaystyle \text{H}^+\) into the solution, which pushes the acid's own equilibrium back toward the undissociated form. This is the common-ion effect.Step $\displaystyle 1$ — get \(\displaystyle K_a\) out of the pKa.\(\displaystyle K_a\) is the acid dissociation constant of acetic acid, and by definition \(\displaystyle pK_a = -\log_{10}K_a\), so\[K_a = 10^{-pK_a} = 10^{-4.74} = 10^{-5}\times10^{0.26} \]Since \(\displaystyle 10^{0.26}\approx 1.82\),\[K_a \approx 1.8\times10^{-5}\ \text{mol L}^{-1} \](pKa is a log scale — you must exponentiate it before using it in ordinary algebra. Never subtract pKa values and treat the result as a concentration.)Step $\displaystyle 2$ — degree of ionization of the acetic acid on its own.Let \(\displaystyle c = 0.05\ \text{mol L}^{-1}\) be the starting concentration of \(\displaystyle \text{CH}_3\text{COOH}\), and \(\displaystyle \alpha\) the fraction of it that ionizes:\[\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]At equilibrium, \(\displaystyle [\text{CH}_3\text{COOH}] = c(1-\alpha)\), \(\displaystyle [\text{CH}_3\text{COO}^-] = c\alpha\), and \(\displaystyle [\text{H}^+] = c\alpha\), so\[K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} = \frac{(c\alpha)(c\alpha)}{c(1-\alpha)} = \frac{c\alpha^2}{1-\alpha} \]\(\displaystyle K_a\) is tiny compared with \(\displaystyle c\), which means \(\displaystyle \alpha\) will come out small, so \(\displaystyle 1-\alpha\approx1\):\[K_a \approx c\alpha^2 \quad\Rightarrow\quad \alpha = \sqrt{\dfrac{K_a}{c}} \]\[\alpha = \sqrt{\frac{1.8\times10^{-5}}{0.05}} = \sqrt{3.6\times10^{-4}} = 1.9\times10^{-2} \](Check the approximation you just made: \(\displaystyle \alpha\approx0.019\), so \(\displaystyle 1-\alpha\approx0.98\) — within $\displaystyle 2$% of $\displaystyle 1$, well inside the precision that a two-significant-figure \(\displaystyle K_a\) can justify.)So on its own, about $\displaystyle 1.9$% of the acetic acid ionizes.Step $\displaystyle 3$ — adding HCl: the common-ion effect.HCl is a strong acid, so it dissociates completely: a "$\displaystyle 0.01$ M HCl" solution already has \(\displaystyle [\text{H}^+] = 0.01\ \text{mol L}^{-1}\) in it before the acetic acid ionizes at all. That \(\displaystyle \text{H}^+\) is the same species the acetic acid equilibrium produces, so by Le Chatelier's principle this extra \(\displaystyle \text{H}^+\) shifts the equilibrium back toward \(\displaystyle \text{CH}_3\text{COOH}\), shrinking the degree of ionization from \(\displaystyle \alpha\) to a smaller value \(\displaystyle \alpha'\).Let \(\displaystyle C\) stand for the HCl concentration. Now \(\displaystyle [\text{CH}_3\text{COOH}] = c(1-\alpha')\), \(\displaystyle [\text{CH}_3\text{COO}^-] = c\alpha'\), and — this is the step that trips people up — \(\displaystyle [\text{H}^+]\) is the total from both sources, \(\displaystyle C + c\alpha'\), not just the acid's own contribution:\[K_a = \frac{c\alpha'\,(C+c\alpha')}{c(1-\alpha')} \]Because \(\displaystyle K_a\) is so small, \(\displaystyle \alpha'\) here will be far smaller than the \(\displaystyle \alpha\) of Step $\displaystyle 2$, so \(\displaystyle c\alpha' \ll C\) and also \(\displaystyle 1-\alpha'\approx1\). Both approximations collapse the expression to\[K_a \approx \alpha' C \quad\Rightarrow\quad \alpha' = \frac{K_a}{C} \](a) \(\displaystyle C = 0.01\ \text{mol L}^{-1}\):\[\alpha' = \frac{1.8\times10^{-5}}{0.01} = 1.8\times10^{-3} \]Check: \(\displaystyle c\alpha' = 0.05 \times 1.8\times10^{-3} = 9\times10^{-5}\ \text{mol L}^{-1}\) — under $\displaystyle 1$% of \(\displaystyle C = 0.01\ \text{mol L}^{-1}\), so \(\displaystyle [\text{H}^+]\approx C\) was a safe approximation.(b) \(\displaystyle C = 0.1\ \text{mol L}^{-1}\):\[\alpha' = \frac{1.8\times10^{-5}}{0.1} = 1.8\times10^{-4} \]Check: \(\displaystyle c\alpha' = 0.05 \times 1.8\times10^{-4} = 9\times10^{-6}\ \text{mol L}^{-1}\) — negligible next to \(\displaystyle C = 0.1\ \text{mol L}^{-1}\), so the approximation holds even better here.Lay the three results side by side: \(\displaystyle 1.9\times10^{-2} \to 1.8\times10^{-3} \to 1.8\times10^{-4}\). Each ten-fold increase in the \(\displaystyle \text{H}^+\) supplied by HCl cuts the acetic acid's own ionization by almost exactly a factor of ten, matching \(\displaystyle \alpha' = K_a/C\) — the signature of common-ion suppression, not a coincidence.Answer: In $\displaystyle 0.05$ M acetic acid alone, \(\displaystyle \alpha \approx 1.9\times10^{-2}\) (about $\displaystyle 1.9$%). Adding $\displaystyle 0.01$ M HCl suppresses this to \(\displaystyle \alpha' \approx 1.8\times10^{-3}\) (about $\displaystyle 0.18$%), and adding $\displaystyle 0.1$ M HCl suppresses it further to \(\displaystyle \alpha' \approx 1.8\times10^{-4}\) (about $\displaystyle 0.018$%) — the degree of dissociation falls sharply as more common \(\displaystyle \text{H}^+\) ion is introduced.

A base's ionization constant \(\displaystyle K_b\) sets how far it splits into ions at a given concentration — and loading the solution with a strong base that supplies the very same product ion (\(\displaystyle \mathrm{OH^-}\)) pushes that split back down. That's the common-ion effect, and it's the part of this problem that catches people off guard.Dimethylamine ionizes in water as\[(\mathrm{CH_3})_2\mathrm{NH} + \mathrm{H_2O} \rightleftharpoons (\mathrm{CH_3})_2\mathrm{NH_2^+} + \mathrm{OH^-} \]Part $\displaystyle 1$ — degree of ionization in plain $\displaystyle 0.02$ M solutionLet \(\displaystyle \alpha\) be the degree of ionization (the fraction of dimethylamine molecules that ionize) and \(\displaystyle C = 0.02\ \mathrm{M}\) the starting concentration. At equilibrium:\[[(\mathrm{CH_3})_2\mathrm{NH}] = C(1-\alpha), \qquad [(\mathrm{CH_3})_2\mathrm{NH_2^+}] = C\alpha, \qquad [\mathrm{OH^-}] = C\alpha \]The base ionization constant is defined as\[K_b = \frac{[(\mathrm{CH_3})_2\mathrm{NH_2^+}][\mathrm{OH^-}]}{[(\mathrm{CH_3})_2\mathrm{NH}]} = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha} \]Aside — this is the step where a weak-electrolyte problem usually gets simplified: when \(\displaystyle \alpha\) turns out to be small, \(\displaystyle 1-\alpha \approx 1\), and the messy quotient collapses to \(\displaystyle K_b \approx C\alpha^2\). Try that first; it's the standard move for this kind of ionization constant.\[K_b \approx C\alpha^2 \quad\Rightarrow\quad \alpha = \sqrt{\frac{K_b}{C}} \]Substituting \(\displaystyle K_b = 5.4\times10^{-4}\) and \(\displaystyle C = 0.02\ \mathrm{M}\):\[\alpha = \sqrt{\frac{5.4\times10^{-4}}{0.02\ \mathrm{M}}} = \sqrt{2.7\times10^{-2}} = 0.164 \]So roughly $\displaystyle 16.4$% of the dimethylamine ionizes when it's on its own in solution.Part $\displaystyle 2$ — ionization once the solution is also $\displaystyle 0.1$ M in NaOHNaOH is a strong base: every formula unit dissociates completely, so it hands the solution \(\displaystyle 0.1\ \mathrm{M}\ \mathrm{OH^-}\) before the dimethylamine ionizes at all. That \(\displaystyle \mathrm{OH^-}\) is a product of the dimethylamine equilibrium too, so by Le Chatelier's principle the extra \(\displaystyle \mathrm{OH^-}\) already present pushes the equilibrium back toward the un-ionized amine — suppressing ionization rather than leaving it unchanged. This is the common-ion effect, and skipping it (treating the two bases as independent) is the error to watch for here.Let \(\displaystyle x\) be the concentration of dimethylamine that now ionizes. At equilibrium:\[[(\mathrm{CH_3})_2\mathrm{NH}] = 0.02\ \mathrm{M}-x, \qquad [(\mathrm{CH_3})_2\mathrm{NH_2^+}] = x, \qquad [\mathrm{OH^-}] = 0.1\ \mathrm{M}+x \]\[K_b = \frac{x(0.1\ \mathrm{M}+x)}{0.02\ \mathrm{M}-x} \]Aside — because the common-ion effect suppresses ionization even further than in Part $\displaystyle 1$, \(\displaystyle x\) will be much smaller than either $\displaystyle 0.02$ M or $\displaystyle 0.1$ M (smaller, in fact, than the \(\displaystyle \alpha\) found in Part $\displaystyle 1$). That makes \(\displaystyle 0.1\ \mathrm{M}+x \approx 0.1\ \mathrm{M}\) and \(\displaystyle 0.02\ \mathrm{M}-x \approx 0.02\ \mathrm{M}\) safe approximations, even though the same kind of approximation was borderline in Part 1.\[K_b \approx \frac{x(0.1\ \mathrm{M})}{0.02\ \mathrm{M}} \]Solving for \(\displaystyle x\):\[x = \frac{K_b \times 0.02\ \mathrm{M}}{0.1\ \mathrm{M}} = \frac{(5.4\times10^{-4})(0.02\ \mathrm{M})}{0.1} = 1.08\times10^{-4}\ \mathrm{M} \]The percentage of dimethylamine ionized is this ionized concentration divided by the original concentration \(\displaystyle C = 0.02\ \mathrm{M}\) — not by $\displaystyle 0.1$ M, which is the NaOH concentration, not the amine's:\[\%\ \text{ionized} = \frac{x}{C}\times 100 = \frac{1.08\times10^{-4}\ \mathrm{M}}{0.02\ \mathrm{M}}\times 100 = 0.54\% \]Aside — that's about $\displaystyle 30$ times smaller than the $\displaystyle 16.4$% found with no NaOH present. \(\displaystyle K_b\) itself hasn't changed (it's fixed at a given temperature); only the equilibrium position has shifted, which is exactly what the common-ion effect predicts.Rounding to match the precision \(\displaystyle K_b = 5.4\times10^{-4}\) (two significant figures) supports: the degree of ionization in plain solution is \(\displaystyle \alpha \approx 0.164\), and the percentage ionized with NaOH present is \(\displaystyle 0.54\%\).Answer: In $\displaystyle 0.02$ M dimethylamine alone, the degree of ionization is \(\displaystyle \alpha \approx 0.164\) (about $\displaystyle 16.4$%). With the solution also $\displaystyle 0.1$ M in NaOH, the common-ion effect suppresses ionization to about $\displaystyle 0.54$%.

pH tells you the power of ten that \(\displaystyle [\mathrm{H^+}]\) sits at — you just have to undo the logarithm.The defining relation is\[\text{pH} = -\log_{10}[\mathrm{H^+}] \]where \(\displaystyle [\mathrm{H^+}]\) is the hydrogen-ion concentration in mol/L (mol L\(\displaystyle ^{-1}\)). Rearranging for the concentration gives\[[\mathrm{H^+}] = 10^{-\text{pH}} \]The step people fumble here is the algebra of the rearrangement itself: from \(\displaystyle \text{pH} = -\log[\mathrm{H^+}]\), multiplying by \(\displaystyle -1\) gives \(\displaystyle -\text{pH} = \log[\mathrm{H^+}]\), and taking the antilog of both sides gives \(\displaystyle [\mathrm{H^+}] = 10^{-\text{pH}}\) — not \(\displaystyle 10^{\text{pH}}\). A larger pH must give a smaller \(\displaystyle [\mathrm{H^+}]\), so the exponent has to stay negative.For each fluid you split \(\displaystyle -\text{pH}\) into a negative integer plus a positive decimal, so the answer comes out as a clean number times a power of ten: \(\displaystyle 10^{-6.83} = 10^{0.17}\times 10^{-7}\), for instance, using \(\displaystyle 10^{0.17}\) as the "clean number."(a) Human muscle fluid, pH = $\displaystyle 6.83$\[[\mathrm{H^+}] = 10^{-6.83} = 10^{0.17}\times 10^{-7}\ \text{mol L}^{-1} \]\(\displaystyle 10^{0.17} = 1.479\), so\[[\mathrm{H^+}] = 1.479\times 10^{-7}\ \text{mol L}^{-1} \approx 1.48\times 10^{-7}\ \text{mol L}^{-1} \](b) Human stomach fluid, pH = $\displaystyle 1.2$\[[\mathrm{H^+}] = 10^{-1.2} = 10^{0.8}\times 10^{-2}\ \text{mol L}^{-1} \]\(\displaystyle 10^{0.8} = 6.310\), so\[[\mathrm{H^+}] = 6.310\times 10^{-2}\ \text{mol L}^{-1} \approx 6.31\times 10^{-2}\ \text{mol L}^{-1} \]The very low pH of stomach fluid is exactly why this concentration is the largest of the four — nearly a million times more concentrated in \(\displaystyle \mathrm{H^+}\) than the blood in part (c).(c) Human blood, pH = $\displaystyle 7.38$\[[\mathrm{H^+}] = 10^{-7.38} = 10^{0.62}\times 10^{-8}\ \text{mol L}^{-1} \]\(\displaystyle 10^{0.62} = 4.169\), so\[[\mathrm{H^+}] = 4.169\times 10^{-8}\ \text{mol L}^{-1} \approx 4.17\times 10^{-8}\ \text{mol L}^{-1} \](d) Human saliva, pH = $\displaystyle 6.4$\[[\mathrm{H^+}] = 10^{-6.4} = 10^{0.6}\times 10^{-7}\ \text{mol L}^{-1} \]\(\displaystyle 10^{0.6} = 3.981\), so\[[\mathrm{H^+}] = 3.981\times 10^{-7}\ \text{mol L}^{-1} \approx 3.98\times 10^{-7}\ \text{mol L}^{-1} \]Each pH value was given to two decimal places, so each \(\displaystyle [\mathrm{H^+}]\) is rounded to three significant figures — carrying more digits would claim precision the data doesn't have.Answer: (a) \(\displaystyle [\mathrm{H^+}] \approx 1.48\times 10^{-7}\) mol L\(\displaystyle ^{-1}\); (b) \(\displaystyle [\mathrm{H^+}] \approx 6.31\times 10^{-2}\) mol L\(\displaystyle ^{-1}\); (c) \(\displaystyle [\mathrm{H^+}] \approx 4.17\times 10^{-8}\) mol L\(\displaystyle ^{-1}\); (d) \(\displaystyle [\mathrm{H^+}] \approx 3.98\times 10^{-7}\) mol L\(\displaystyle ^{-1}\)

pH is defined as \(\displaystyle \text{pH} = -\log_{10}[\text{H}^+] \), so reversing it gives \(\displaystyle [\text{H}^+] = 10^{-\text{pH}} \) — the hydrogen ion concentration is ten raised to the negative of the pH value. Do this for each liquid.To evaluate \(\displaystyle 10^{-\text{pH}}\) by hand, split the exponent into a whole-number part and a decimal part, because \(\displaystyle 10^{-n}\) (n a whole number) is easy, and \(\displaystyle 10^{0.x}\) is a small factor you can read off (or compute) separately.Milk, pH = $\displaystyle 6.8$Write \(\displaystyle -6.8 = 0.2 - 7\), so \[[\text{H}^+] = 10^{-6.8} = 10^{0.2} \times 10^{-7} = 1.585 \times 10^{-7}\ \text{mol L}^{-1} \] Rounded to two significant figures (matching the one-decimal precision of the pH data): \(\displaystyle 1.6 \times 10^{-7}\ \text{mol L}^{-1}\).Black coffee, pH = $\displaystyle 5.0$Here the exponent is already a whole number, so no splitting is needed: \[[\text{H}^+] = 10^{-5.0} = 1.0 \times 10^{-5}\ \text{mol L}^{-1} \]Tomato juice, pH = $\displaystyle 4.2$Write \(\displaystyle -4.2 = 0.8 - 5\), so \[[\text{H}^+] = 10^{-4.2} = 10^{0.8} \times 10^{-5} = 6.310 \times 10^{-5}\ \text{mol L}^{-1} \] Rounded: \(\displaystyle 6.3 \times 10^{-5}\ \text{mol L}^{-1}\).Lemon juice, pH = $\displaystyle 2.2$Write \(\displaystyle -2.2 = 0.8 - 3\), so \[[\text{H}^+] = 10^{-2.2} = 10^{0.8} \times 10^{-3} = 6.310 \times 10^{-3}\ \text{mol L}^{-1} \] Rounded: \(\displaystyle 6.3 \times 10^{-3}\ \text{mol L}^{-1}\).Egg white, pH = $\displaystyle 7.8$Write \(\displaystyle -7.8 = 0.2 - 8\), so \[[\text{H}^+] = 10^{-7.8} = 10^{0.2} \times 10^{-8} = 1.585 \times 10^{-8}\ \text{mol L}^{-1} \] Rounded: \(\displaystyle 1.6 \times 10^{-8}\ \text{mol L}^{-1}\).A step people get wrong here: the sign in \(\displaystyle [\text{H}^+] = 10^{-\text{pH}} \) is easy to drop, giving a huge concentration instead of a tiny one — a bigger pH number must always turn into a smaller \(\displaystyle [\text{H}^+]\), since pH and \(\displaystyle [\text{H}^+]\) move in opposite directions (milk, the least acidic of the five, correctly comes out with the smallest \(\displaystyle [\text{H}^+]\), and lemon juice, the most acidic, with the largest).Answer: [H⁺] = $\displaystyle 1.6$ × $\displaystyle 10$⁻⁷ mol L⁻¹ (milk), $\displaystyle 1.0$ × $\displaystyle 10$⁻⁵ mol L⁻¹ (black coffee), $\displaystyle 6.3$ × $\displaystyle 10$⁻⁵ mol L⁻¹ (tomato juice), $\displaystyle 6.3$ × $\displaystyle 10$⁻³ mol L⁻¹ (lemon juice), $\displaystyle 1.6$ × $\displaystyle 10$⁻⁸ mol L⁻¹ (egg white)

KOH is a strong base — it dissociates completely, so the concentration of \(\displaystyle \text{K}^+ \) and \(\displaystyle \text{OH}^- \) equals the concentration of KOH itself, not some equilibrium fraction of it.Step $\displaystyle 1$ — moles of KOH.Molar mass of KOH: \(\displaystyle M = 39 + 16 + 1 = 56\ \text{g mol}^{-1} \) (K = $\displaystyle 39$, O = $\displaystyle 16$, H = $\displaystyle 1$).Moles, \(\displaystyle n = \dfrac{\text{mass}}{\text{molar mass}} \), where mass is the mass of solute ($\displaystyle 0.561$ g) and molar mass is $\displaystyle 56$ g mol⁻¹:\[n(\text{KOH}) = \frac{0.561\ \text{g}}{56\ \text{g mol}^{-1}} = 1.002 \times 10^{-2}\ \text{mol} \]Step $\displaystyle 2$ — molar concentration of the KOH solution.Molarity \(\displaystyle c = \dfrac{n}{V} \), where \(\displaystyle n\) is moles of solute and \(\displaystyle V\) is the volume of the solution in litres.Aside: the volume that goes in the denominator is the $\displaystyle 200$ mL of final solution stated in the problem — not the volume of water used to make it. Mixing up "volume of solvent" with "volume of solution" is the error that most often derails this kind of question.\[V = 200\ \text{mL} = 0.200\ \text{L} \]\[c(\text{KOH}) = \frac{1.002 \times 10^{-2}\ \text{mol}}{0.200\ \text{L}} = 5.01 \times 10^{-2}\ \text{mol L}^{-1} \]Step $\displaystyle 3$ — concentrations of \(\displaystyle \text{K}^+ \) and \(\displaystyle \text{OH}^- \).KOH dissociates completely, one mole of solid giving one mole each of the ions:\[\text{KOH} \longrightarrow \text{K}^+ + \text{OH}^- \]Because the stoichiometry is $\displaystyle 1$ : $\displaystyle 1$ : $\displaystyle 1$ and dissociation is complete (strong base — no equilibrium arrow needed here), each ion's concentration equals the concentration of KOH found above:\[[\text{K}^+] = [\text{OH}^-] = 5.01 \times 10^{-2}\ \text{mol L}^{-1} \]Step $\displaystyle 4$ — \(\displaystyle [\text{H}^+] \) from the ionic product of water.Ionic product of water: \(\displaystyle K_w = [\text{H}^+][\text{OH}^-] \), with \(\displaystyle K_w = 1.0 \times 10^{-14}\ \text{mol}^2\text{L}^{-2} \) at $\displaystyle 298$ K — this holds in any aqueous solution, acidic or basic, at this temperature.\[[\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.0 \times 10^{-14}}{5.01 \times 10^{-2}} = 2.00 \times 10^{-13}\ \text{mol L}^{-1} \]Step $\displaystyle 5$ — pH.Definition: \(\displaystyle \text{pH} = -\log_{10}[\text{H}^+] \), where \(\displaystyle [\text{H}^+] \) is in mol L⁻¹.\[\text{pH} = -\log\left(2.00 \times 10^{-13}\right) = 13 - \log(2.00) = 13 - 0.30 = 12.70 \]Aside — a quick check on this route: it is easy to forget the minus sign and to trip over subtracting a negative power of ten; going the other way through pOH catches such slips. \(\displaystyle \text{pOH} = -\log[\text{OH}^-] = -\log(5.01\times10^{-2}) = 1.30 \), and since \(\displaystyle \text{pH} + \text{pOH} = 14.00 \) at $\displaystyle 298$ K, \(\displaystyle \text{pH} = 14.00 - 1.30 = 12.70 \) — the same value, confirming the result.Data are given to $\displaystyle 3$ significant figures ($\displaystyle 0.561$ g), so the concentrations are reported to $\displaystyle 3$ significant figures and pH, being a logarithmic quantity, to $\displaystyle 2$ decimal places.Answer: \(\displaystyle [\text{K}^+] = [\text{OH}^-] = 5.01 \times 10^{-2}\ \text{mol L}^{-1} \) (≈ $\displaystyle 0.05$ M); \(\displaystyle [\text{H}^+] = 2.00 \times 10^{-13}\ \text{mol L}^{-1} \); pH = $\displaystyle 12.70$

The solubility is given as grams of compound per litre of solution — divide by molar mass to get molarity, the same way you always convert g/L to mol/L.Step $\displaystyle 1$: Molar mass of \(\displaystyle \mathrm{Sr(OH)_2} \)Using atomic masses \(\displaystyle M_{Sr} = 87.6\ \text{g/mol} \), \(\displaystyle M_O = 16.0\ \text{g/mol} \), \(\displaystyle M_H = 1.0\ \text{g/mol} \):\[M\big(\mathrm{Sr(OH)_2}\big) = 87.6 + 2(16.0 + 1.0) = 87.6 + 34.0 = 121.6\ \text{g/mol} \]Step $\displaystyle 2$: Convert solubility to molarityMolarity \(\displaystyle c \) — moles of solute per litre of solution (the $\displaystyle 19.23$ g is already given per litre of solution, so no further correction for solvent volume is needed) — is\[c = \frac{\text{mass of solute per litre of solution}}{\text{molar mass}} = \frac{19.23\ \text{g/L}}{121.6\ \text{g/mol}} = 0.1581\ \text{mol/L} \]This \(\displaystyle c \) is the concentration of dissolved \(\displaystyle \mathrm{Sr(OH)_2} \) as a whole, before it splits into ions.Step $\displaystyle 3$: Dissociation gives the ion concentrations\(\displaystyle \mathrm{Sr(OH)_2} \) is a strong base, so it dissociates completely in water:\[\mathrm{Sr(OH)_2(aq)} \;\rightarrow\; \mathrm{Sr^{2+}(aq)} \;+\; 2\,\mathrm{OH^-(aq)} \]One formula unit releases one \(\displaystyle \mathrm{Sr^{2+}} \) but two \(\displaystyle \mathrm{OH^-} \) — missing that factor of $\displaystyle 2$ is the step most people get wrong here.\[[\mathrm{Sr^{2+}}] = c = 0.1581\ \text{mol/L} \approx 0.158\ \text{M} \]\[[\mathrm{OH^-}] = 2c = 2 \times 0.1581\ \text{mol/L} = 0.3163\ \text{mol/L} \approx 0.316\ \text{M} \]Step $\displaystyle 4$: pH from \(\displaystyle [\mathrm{OH^-}] \)The formula \(\displaystyle \text{pOH} = -\log_{10}[\mathrm{OH^-}] \) must use the hydroxide-ion concentration, not the concentration of dissolved \(\displaystyle \mathrm{Sr(OH)_2} \) — plugging in \(\displaystyle c \) instead of \(\displaystyle 2c \) is the other common slip.\[\text{pOH} = -\log_{10}(0.3163) = 0.500 \]At $\displaystyle 298$ K, \(\displaystyle \text{pH} + \text{pOH} = 14 \), so\[\text{pH} = 14 - 0.500 = 13.50 \]Answer: \(\displaystyle [\mathrm{Sr^{2+}}] \approx 0.158\ \text{M}\), \(\displaystyle [\mathrm{OH^-}] \approx 0.316\ \text{M}\), and pH = $\displaystyle 13.5$

The acid barely ionizes on its own, but adding H⁺ from a strong acid pushes the equilibrium back — that's Le Chatelier acting through a common ion.Step $\displaystyle 1$: Set up the ionization equilibrium for the pure acid.Propanoic acid ionizes as \[\text{C}_2\text{H}_5\text{COOH} \rightleftharpoons \text{C}_2\text{H}_5\text{COO}^- + \text{H}^+ \]Let \(\displaystyle C = 0.05\ \text{M}\) be the initial concentration and \(\displaystyle \alpha\) the degree of ionization (the fraction of acid molecules that have ionized). At equilibrium:

	\(\displaystyle \text{C}_2\text{H}_5\text{COOH}\)	\(\displaystyle \text{C}_2\text{H}_5\text{COO}^-\)	\(\displaystyle \text{H}^+\)
Initial	\(\displaystyle C\)	$\displaystyle 0$	$\displaystyle 0$
Equilibrium	\(\displaystyle C(1-\alpha)\)	\(\displaystyle C\alpha\)	\(\displaystyle C\alpha\)

The ionization constant is \[K_a = \frac{[\text{C}_2\text{H}_5\text{COO}^-][\text{H}^+]}{[\text{C}_2\text{H}_5\text{COOH}]} = \frac{C\alpha \cdot C\alpha}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha} \]Since \(\displaystyle K_a\) is small (of order \(\displaystyle 10^{-5}\)), \(\displaystyle \alpha \ll 1\), so \(\displaystyle 1-\alpha \approx 1\). This is the approximation everyone reaches for — but it is only valid when the resulting \(\displaystyle \alpha\) turns out to be small, so check it at the end. With that simplification, \[K_a \approx C\alpha^2 \quad\Rightarrow\quad \alpha = \sqrt{\frac{K_a}{C}} \]Step $\displaystyle 2$: Substitute the numbers.\[\alpha = \sqrt{\frac{1.32\times10^{-5}}{0.05}} = \sqrt{2.64\times10^{-4}} = 1.625\times10^{-2} \]Check the approximation: \(\displaystyle \alpha = 0.01625\), which is far below \(\displaystyle 0.05\) (the usual $\displaystyle 5$% cutoff), so treating \(\displaystyle 1-\alpha \approx 1\) was justified.Step $\displaystyle 3$: Find \(\displaystyle [\text{H}^+]\) and then pH.\[[\text{H}^+] = C\alpha = 0.05\ \text{M} \times 1.625\times10^{-2} = 8.12\times10^{-4}\ \text{M} \]\[\text{pH} = -\log[\text{H}^+] = -\log(8.12\times10^{-4}) \]Splitting the log: \(\displaystyle -\log(8.12\times10^{-4}) = 4 - \log(8.12) = 4 - 0.910 = 3.09\)Step $\displaystyle 4$: Now add $\displaystyle 0.01$ M HCl — the common-ion effect.HCl is a strong acid, so it dissociates completely and dumps \(\displaystyle 0.01\ \text{M}\ \text{H}^+\) into the solution before the propanoic acid equilibrium is even considered. This extra \(\displaystyle \text{H}^+\) is a common ion (propanoic acid also produces \(\displaystyle \text{H}^+\)), and by Le Chatelier's principle it suppresses the acid's own ionization — the acid's degree of ionization must now be smaller than \(\displaystyle 1.625\times10^{-2}\).Let \(\displaystyle \alpha'\) be the new degree of ionization of propanoic acid (still starting from \(\displaystyle C = 0.05\ \text{M}\)). The equilibrium concentrations are now:

	\(\displaystyle \text{C}_2\text{H}_5\text{COOH}\)	\(\displaystyle \text{C}_2\text{H}_5\text{COO}^-\)	\(\displaystyle \text{H}^+\)
Initial	\(\displaystyle 0.05\)	$\displaystyle 0$	\(\displaystyle 0.01\) (from HCl)
Equilibrium	\(\displaystyle 0.05(1-\alpha')\)	\(\displaystyle 0.05\,\alpha'\)	\(\displaystyle 0.01 + 0.05\,\alpha'\)

\[K_a = \frac{(0.05\,\alpha')(0.01 + 0.05\,\alpha')}{0.05(1-\alpha')} \]The step people skip: you cannot drop the \(\displaystyle 0.05\alpha'\) sitting next to the \(\displaystyle 0.01\) until you've confirmed it really is negligible next to it — not just small in absolute terms. Since the common-ion effect makes \(\displaystyle \alpha'\) very small here, expect \(\displaystyle 0.05\,\alpha' \ll 0.01\) and \(\displaystyle \alpha' \ll 1\), and verify after solving. With \(\displaystyle 1-\alpha' \approx 1\) and \(\displaystyle 0.01+0.05\alpha' \approx 0.01\):\[K_a \approx \frac{(0.05\,\alpha')(0.01)}{0.05} = 0.01\,\alpha' \]\[\alpha' = \frac{K_a}{0.01} = \frac{1.32\times10^{-5}}{1\times10^{-2}} = 1.32\times10^{-3} \]Step $\displaystyle 5$: Verify the approximation.\(\displaystyle 0.05\,\alpha' = 0.05 \times 1.32\times10^{-3} = 6.6\times10^{-5}\), which is indeed negligible next to \(\displaystyle 0.01\) (less than $\displaystyle 1$% of it), and \(\displaystyle \alpha' = 1.32\times10^{-3} \ll 1\). Both simplifications hold.Comparing the two results: \(\displaystyle \alpha\) fell from \(\displaystyle 1.625\times10^{-2}\) (about $\displaystyle 1.6$%) to \(\displaystyle 1.32\times10^{-3}\) (about $\displaystyle 0.13$%) once the common \(\displaystyle \text{H}^+\) from HCl was added — roughly a $\displaystyle 12$-fold suppression, exactly the direction Le Chatelier's principle predicts.Answer: Degree of ionization of $\displaystyle 0.05$ M propanoic acid \(\displaystyle \alpha = 1.62\times10^{-2}\) ($\displaystyle 1.6$%), with pH \(\displaystyle = 3.09\). In the presence of $\displaystyle 0.01$ M HCl, the degree of ionization drops to \(\displaystyle \alpha' = 1.32\times10^{-3}\) ($\displaystyle 0.13$%).

A weak acid's \(\displaystyle K_a\) is built from the concentrations left at equilibrium — and the acid that ionized is no longer acid. That single bookkeeping step is the whole question here.Cyanic acid is a weak monoprotic acid, so it ionizes only partly:\[\mathrm{HCNO} \rightleftharpoons \mathrm{H^+} + \mathrm{CNO^-} \]Step $\displaystyle 1$ — get \(\displaystyle [\mathrm{H^+}]\) from the pH.The definition of pH is \(\displaystyle \mathrm{pH} = -\log_{10}[\mathrm{H^+}]\), where \(\displaystyle [\mathrm{H^+}]\) is the hydrogen-ion concentration in mol L\(\displaystyle ^{-1}\). Inverting it:\[[\mathrm{H^+}] = 10^{-\mathrm{pH}} = 10^{-2.34} = 4.571 \times 10^{-3}\ \text{mol L}^{-1} \]Aside: pH is a logarithm, so you invert it with a power of ten, never by dividing. And \(\displaystyle 10^{-2.34}\) is not \(\displaystyle 2.34 \times 10^{-3}\) — write it as \(\displaystyle 10^{0.66} \times 10^{-3} = 4.571 \times 10^{-3}\).Step $\displaystyle 2$ — build the equilibrium table.Let \(\displaystyle C = 0.1\) mol L\(\displaystyle ^{-1}\) be the concentration of acid taken initially, and let \(\displaystyle x\) be the amount per litre that ionizes.

	\(\displaystyle \mathrm{HCNO}\)	\(\displaystyle \mathrm{H^+}\)	\(\displaystyle \mathrm{CNO^-}\)
Initial	\(\displaystyle C = 0.1\)	\(\displaystyle 0\)	\(\displaystyle 0\)
Change	\(\displaystyle -x\)	\(\displaystyle +x\)	\(\displaystyle +x\)
Equilibrium	\(\displaystyle C - x\)	\(\displaystyle x\)	\(\displaystyle x\)

Each molecule that ionizes gives one \(\displaystyle \mathrm{H^+}\) and one \(\displaystyle \mathrm{CNO^-}\), so \(\displaystyle x = [\mathrm{H^+}] = 4.571 \times 10^{-3}\) mol L\(\displaystyle ^{-1}\). (The \(\displaystyle \mathrm{H^+}\) from water, about \(\displaystyle 10^{-7}\) mol L\(\displaystyle ^{-1}\), is smaller by a factor of \(\displaystyle 4\times10^{4}\) and is safely ignored.)Equilibrium concentrations:\[[\mathrm{H^+}] = [\mathrm{CNO^-}] = 4.571 \times 10^{-3}\ \text{mol L}^{-1} \] \[[\mathrm{HCNO}] = C - x = 0.1 - 0.004571 = 0.09543\ \text{mol L}^{-1} \]Aside — this is the step people lose the mark on: the denominator is the acid remaining, \(\displaystyle C - x\), not the \(\displaystyle 0.1\) you started with. What ionized has left the HCNO column and moved into the ion columns.Step $\displaystyle 3$ — the ionization constant.\[K_a = \frac{[\mathrm{H^+}][\mathrm{CNO^-}]}{[\mathrm{HCNO}]} = \frac{x^2}{C-x} \]Substituting:\[K_a = \frac{(4.571 \times 10^{-3})^2}{0.09543} = \frac{2.0893 \times 10^{-5}}{0.09543} = 2.189 \times 10^{-4} \]The pH is given to two decimal places, which fixes \(\displaystyle [\mathrm{H^+}]\) to about three significant figures; \(\displaystyle K_a\) depends on the square of it, so quote three figures at most:\[K_a \approx 2.19 \times 10^{-4} \]Equivalently \(\displaystyle \mathrm{p}K_a = -\log K_a = 3.66\).Aside: \(\displaystyle K_a\) is dimensionless as it is written here, because the concentrations are ratios to the standard state of $\displaystyle 1$ mol L\(\displaystyle ^{-1}\). You will also see it quoted with mol L\(\displaystyle ^{-1}\); either convention is accepted, but do not mix them.Step $\displaystyle 4$ — the degree of ionization.The degree of ionization \(\displaystyle \alpha\) is the fraction of the acid taken that has ionized:\[\alpha = \frac{\text{amount ionized per litre}}{\text{amount taken per litre}} = \frac{x}{C} \]\[\alpha = \frac{4.571 \times 10^{-3}}{0.1} = 4.57 \times 10^{-2} = 0.0457 \]That is \(\displaystyle 4.57\%\) of the cyanic acid ionized — about one molecule in twenty-two.Aside: \(\displaystyle \alpha\) is a fraction, so it has no unit and can never exceed 1. If you get \(\displaystyle \alpha > 1\), you have divided by \(\displaystyle x\) instead of by \(\displaystyle C\).A note on the two values of \(\displaystyle K_a\) you will see.NCERT prints \(\displaystyle K_a = 2.09 \times 10^{-4}\) for this question. That comes from the standard weak-acid shortcut, in which \(\displaystyle x\) is dropped from the denominator because it is small compared with \(\displaystyle C\):\[K_a \approx \frac{x^2}{C} = \frac{2.0893 \times 10^{-5}}{0.1} = 2.09 \times 10^{-4} \]This is a deliberate approximation, not a misprint, and it is the one usually allowed when \(\displaystyle \alpha < 5\%\) — here \(\displaystyle \alpha = 4.57\%\), just inside that limit. But notice how thin the margin is: dropping \(\displaystyle x\) shifts \(\displaystyle K_a\) by \(\displaystyle 4.8\%\), which is larger than the uncertainty the given pH carries (about \(\displaystyle 2\%\) in \(\displaystyle K_a\)). Since the exact denominator costs one subtraction, use \(\displaystyle C - x\) and report \(\displaystyle 2.19 \times 10^{-4}\); if your answer key or examiner works with the approximate form, \(\displaystyle 2.09 \times 10^{-4}\) is the number they expect, and both rest on the same \(\displaystyle [\mathrm{H^+}]\). The degree of ionization, \(\displaystyle 0.0457\), is identical either way — it never depended on that choice.Answer: \(\displaystyle K_a = 2.19 \times 10^{-4}\) (three significant figures; \(\displaystyle 2.09 \times 10^{-4}\) if the denominator is approximated as \(\displaystyle 0.1\) M, as NCERT does), and the degree of ionization \(\displaystyle \alpha = 4.57 \times 10^{-2}\), i.e. \(\displaystyle 4.57\%\).