A2 Chemistry 9701: pH of Strong Acids and Bases Using Kw

At $298\ \text{K}$ , the ionic product of water is $K_w = [\text{H}^+][\text{OH}^-] = 1.00\times10^{-14}\ \text{mol}^2\ \text{dm}^{-6}$ .

(a) Hydrochloric acid, $\text{HCl}$ , is a strong monobasic acid that dissociates completely in water. Calculate the pH of a $0.0250\ \text{mol dm}^{-3}$ solution of $\text{HCl}$ at $298\ \text{K}$ . [2]

(b) Sodium hydroxide, $\text{NaOH}$ , is a strong base that dissociates completely in water. Calculate the pH of a $0.0250\ \text{mol dm}^{-3}$ solution of $\text{NaOH}$ at $298\ \text{K}$ , using $K_w$ . [3]

(c) The dissociation of water, $\text{H}_2\text{O(l)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)}$ , is endothermic. State and explain what happens to the value of $K_w$ , and to the pH of pure water, as the temperature is raised above $298\ \text{K}$ . Explain why pure water remains neutral at the higher temperature even though its pH is no longer $7.00$ . [2]

Part (a): pH of the strong acid

$\text{HCl}$ is a strong monobasic acid, so it dissociates completely: $\text{HCl(aq)} \rightarrow \text{H}^+\text{(aq)} + \text{Cl}^-\text{(aq)}$

Since the stoichiometry is $1:1$ , $[\text{H}^+]$ equals the full stated concentration of the acid: $[\text{H}^+] = 0.0250\ \text{mol dm}^{-3}$

$\text{pH} = -\log(0.0250) = 1.602\ \Rightarrow\ \text{pH}=1.60\ \text{(2 d.p.)}$

(Check: $10^{-1.60}=0.0251$ , close to $0.0250$ within rounding. Consistent.)

Part (b): pH of the strong base

$\text{NaOH}$ is a strong base, so it also dissociates completely: $\text{NaOH(aq)} \rightarrow \text{Na}^+\text{(aq)} + \text{OH}^-\text{(aq)}$

Since the stoichiometry is $1:1$ , $[\text{OH}^-]$ equals the full stated concentration: $[\text{OH}^-] = 0.0250\ \text{mol dm}^{-3}$

Rearranging $K_w=[\text{H}^+][\text{OH}^-]$ for $[\text{H}^+]$ : $[\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.00\times10^{-14}}{0.0250} = 4.00\times10^{-13}\ \text{mol dm}^{-3}$

$\text{pH} = -\log(4.00\times10^{-13}) = 12.398\ \Rightarrow\ \text{pH}=12.40\ \text{(2 d.p.)}$

(Check by the pOH route: $\text{pOH}=-\log(0.0250)=1.60$ , and $\text{pH}=14.00-\text{pOH}=14.00-1.60=12.40$ . The two methods agree.)

Part (c): Effect of temperature on Kw and the pH of pure water

Since the forward dissociation of water is endothermic, raising the temperature shifts the position of this equilibrium to the right (Le Chatelier’s principle), increasing both $[\text{H}^+]$ and $[\text{OH}^-]$ . Because $K_w=[\text{H}^+][\text{OH}^-]$ , this means $K_w$ increases above $1.00\times10^{-14}\ \text{mol}^2\ \text{dm}^{-6}$ as temperature rises.

Since $[\text{H}^+]$ increases, $\text{pH}=-\log[\text{H}^+]$ decreases, falling below $7.00$ .

However, water always dissociates in a $1:1$ ratio, so $[\text{H}^+]=[\text{OH}^-]$ at any temperature. Neutrality is defined by this equality, not by a fixed pH value of $7.00$ , so pure water remains neutral at the higher temperature, even though its pH is now below $7.00$ .

Final answers

(a) $\text{pH}=1.60$
(b) $\text{pH}=12.40$
(c) $K_w$ increases and the pH of pure water falls below $7.00$ ; water is still neutral because $[\text{H}^+]=[\text{OH}^-]$ regardless of temperature