Exercises — Ionic product of water Kw = 10⁻¹⁴ at 25 °C

The autoionization ${\text{H}}_2\text{O}\rightleftharpoons {\text{H}}^{+}+{\text{OH}}^{-}$ has equilibrium expression (after absorbing the constant $[{\text{H}}_2\text{O}{]}^2$): $K_w=[{\text{H}}^{+}][{\text{OH}}^{-}]=1.0\times 10^{-14}$ What we did: simply recalled the boxed rule. Nothing to compute — this is pure recognition.

Smaller. Acidic means $[{\text{H}}^{+}]>10^{-7}$. Since the product is fixed at $10^{-14}$, if one factor rises above $10^{-7}$ the other must sink below $10^{-7}$. That is the seesaw: the two concentrations trade off around the neutral value $10^{-7}$.

Rearrange the master rule to isolate the unknown: $[{\text{OH}}^{-}]=\frac{K_w}{[{\text{H}}^{+}]}=\frac{1.0\times 10^{-14}}{2.0\times 10^{-4}}=5.0\times 10^{-11}\text{M}$ Why divide? The product of the two is fixed. Knowing one factor, the other is the fixed product divided by it. Check: $(2.0\times 10^{-4})(5.0\times 10^{-11})=10^{-14}$ ✓. Since $[{\text{H}}^{+}]>[{\text{OH}}^{-}]$, the solution is acidic — consistent.

Use the additive rule: $\text{pOH}=14-\text{pH}=14-5.2=8.8$ Then undo the $-\log$ to get the concentration. "$\text{pOH}=8.8$" means $-{\log }_{10}[{\text{OH}}^{-}]=8.8$, so: $[{\text{OH}}^{-}]=10^{-8.8}=1.6\times 10^{-9}\text{M}$ Why $14-\text{pH}$? Because $\text{pH}+\text{pOH}=14$ is just the log-form of the seesaw. Adding instead of multiplying is the whole point of using p-notation.

Full dissociation means $[{\text{OH}}^{-}]=0.0010=1.0\times 10^{-3}$ M. $\text{pOH}=-{\log }_{10}(10^{-3})=3,\text{pH}=14-3=11$ Why start with pOH? A base hands us $[{\text{OH}}^{-}]$ directly, so it is natural to compute pOH first, then flip to pH with the 14-rule.

In pure water every ${\text{H}}^{+}$ is born alongside one ${\text{OH}}^{-}$, so $[{\text{H}}^{+}]=[{\text{OH}}^{-}]=x$: $x^2=9.6\times 10^{-14}\Rightarrow x=\sqrt{9.6\times 10^{-14}}=3.1\times 10^{-7}\text{M}$ $\text{pH}=-{\log }_{10}(3.1\times 10^{-7})=6.51$ Not acidic. Neutrality is defined by $[{\text{H}}^{+}]=[{\text{OH}}^{-}]$, which still holds here. pH 6.51 is neutral at 60 °C — the neutral pH shifts down because heating drives the endothermic autoionization forward, making more of both ions.

Naïve answer: $\text{pH}=-\log (10^{-8})=8$ — an acid with pH 8? Impossible. The flaw: at this dilution the water's own $10^{-7}$ M of ${\text{H}}^{+}$ is larger than the acid's contribution, so we cannot ignore water. Charge balance and $K_w$ give: $[{\text{H}}^{+}]=\frac{c+\sqrt{c^2+4K_w}}{2},c=1.0\times 10^{-8}$ $[{\text{H}}^{+}]=\frac{10^{-8}+\sqrt{(10^{-8}{)}^2+4(10^{-14})}}{2}=1.05\times 10^{-7}\text{M}$ $\text{pH}=-{\log }_{10}(1.05\times 10^{-7})=6.98$ Just below 7 — faintly acidic, as an acid must be. The acid nudges the water's $10^{-7}$ up ever so slightly.

A conjugate acid–base pair obeys $K_a\times K_b=K_w$ — a direct child of the water equilibrium. So: $K_b=\frac{K_w}{K_a}=\frac{1.0\times 10^{-14}}{5.6\times 10^{-10}}=1.8\times 10^{-5}$ Why does $K_a{K}_b=K_w$? Adding the acid-ionization and base-ionization equations cancels everything except ${\text{H}}_2\text{O}\rightleftharpoons {\text{H}}^{+}+{\text{OH}}^{-}$; multiplying constants of added reactions gives $K_w$.

Count moles of each (millimoles = mL × M): $n_{{\text{H}}^{+}}=50.0\times 0.10=5.0\text{mmol},n_{{\text{OH}}^{-}}=30.0\times 0.10=3.0\text{mmol}$ They neutralise 1:1, leaving excess acid: $n_{{\text{H}}^{+},\text{excess}}=5.0-3.0=2.0\text{mmol}$ Total volume $=50.0+30.0=80.0$ mL: $[{\text{H}}^{+}]=\frac{2.0\text{mmol}}{80.0\text{mL}}=0.025\text{M}$ $\text{pH}=-{\log }_{10}(0.025)=1.60$ Why subtract then divide by total volume? Neutralisation removes matched pairs; the survivors spread across the combined volume, diluting them.

Pure water: $[{\text{H}}^{+}]=[{\text{OH}}^{-}]=x$, so $x=\sqrt{1.14\times 10^{-15}}=3.38\times 10^{-8}\text{M}$ $\text{pH}=-{\log }_{10}(3.38\times 10^{-8})=7.47$ Above 7, yet still neutral. Cooling removes heat; since autoionization is endothermic, Le Chatelier pulls the equilibrium backward, recombining ions. Fewer ions → smaller $[{\text{H}}^{+}]$ → higher pH. Neutrality still means $[{\text{H}}^{+}]=[{\text{OH}}^{-}]$, only the numerical pH of that balance point has drifted upward.

(a) $\text{pH}=-{\log }_{10}(4.0\times 10^{-9})=8.40$ (b) $[{\text{OH}}^{-}]=\frac{10^{-14}}{4.0\times 10^{-9}}=2.5\times 10^{-6}\text{M}$ (c) $\text{pOH}=-{\log }_{10}(2.5\times 10^{-6})=5.60$ (d) $\text{pH}+\text{pOH}=8.40+5.60=14.00$ ✓ What this shows: the four quantities are one fact wearing four costumes. Knowing any one at 25 °C fixes the other three through $K_w$.

In pure water each water-derived ${\text{H}}^{+}$ equals $1.0\times 10^{-7}$ M. With external hydroxide flooding in: $[{\text{H}}^{+}]=\frac{K_w}{[{\text{OH}}^{-}]}=\frac{10^{-14}}{0.10}=1.0\times 10^{-13}\text{M}$ Every ${\text{H}}^{+}$ now comes from water (NaOH adds only ${\text{OH}}^{-}$), so water's ionization dropped from $10^{-7}$ to $10^{-13}$: $\text{factor}=\frac{10^{-7}}{10^{-13}}=10^6$ Interpretation: the extra ${\text{OH}}^{-}$ (a common ion) drives ${\text{H}}^{+}+{\text{OH}}^{-}\to {\text{H}}_2\text{O}$ backward, suppressing water's self-ionization a million-fold — yet the product $[{\text{H}}^{+}][{\text{OH}}^{-}]$ still equals $10^{-14}$.

The master rule at 25 °C is $K_w=$ ==$[{\text{H}}^{+}][{\text{OH}}^{-}]=1.0\times 10^{-14}$. In log form: $\text{pH}+\text{pOH}=14$. For a conjugate pair: $K_a\times K_b=K_w$. Kw depends only on temperature== (it rises when heated because autoionization is endothermic).