Question 14.9: Calculating Ka and pKa for a Weak Acid from the pH of the So......

Calculating KaK_a and pKaK_a for a Weak Acid from the pH of the Solution

The pH of 0.250 M HF is 2.036. What are the values of KaK_a and pKaK_a for hydrofluoric acid?

STRATEGY

First, write the balanced equation for the dissociation equilibrium. Then, define x as the concentration of HF that dissociates and make the usual table under the balanced equation (Steps 1 and 2 of Figure 13.6 on page 522). Because x equals the H3O+H_3O^+ concentration, its value can be calculated from the pH. Finally, substitute the equilibrium concentrations into the equilibrium equation to obtain the value of KaK_a and take the negative log of KaK_a to obtain the pKaK_a.

13.6
Step-by-Step
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                 HF(aq)+H2O(l)             H3O+(aq)+F(aq) Initial concentration (M)0.25000 Change (M)x+x+x Equilibrium concentration (M)(0.250x)xx\begin{array}{lccc} & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{HF}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons & \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{H}_3 \mathrm{O}^{+}(a q)+ & \mathrm{F}^{-}(a q) \\\hline \text { Initial concentration }(\mathrm{M}) & 0.250 & \sim 0^* & 0 \\\text { Change }(\mathrm{M}) & -x & +x & +x \\\text { Equilibrium concentration }(\mathrm{M}) & (0.250-x) & x & x\end{array}

*A very small concentration of H3O+H_3O^+ is present initially because of the dissociation of water.

We can calculate the value of x from the pH:

x=[H3O+]=antilog(pH)=10pH=102.036=9.20×103Mx=\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\operatorname{antilog}(-\mathrm{pH})=10^{-\mathrm{pH}}=10^{-2.036}=9.20 \times 10^{-3} \mathrm{M}

The other equilibrium concentrations are

[F]=x=9.20×103M[HF]=0.250x=0.2500.00920=0.241M\begin{aligned}& {\left[\mathrm{F}^{-}\right]=x=9.20 \times 10^{-3} \mathrm{M}} \\& {[\mathrm{HF}]=0.250-x=0.250-0.00920=0.241 \mathrm{M}}\end{aligned}

Substituting these concentrations into the equilibrium equation gives the value of KaK_a:

Ka=[H3O+][F][HF]=(x)(x)(0.250x)=(9.20×103)(9.20×103)0.241=3.51×104pKa=logKa=log(3.51×104)=3.455K_{\mathrm{a}}=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{F}^{-}\right]}{[\mathrm{HF}]}=\frac{(x)(x)}{(0.250-x)}=\frac{\left(9.20 \times 10^{-3}\right)\left(9.20 \times 10^{-3}\right)}{0.241}=3.51 \times 10^{-4} \\ \mathrm{p} K_{\mathrm{a}}=-\log K_{\mathrm{a}}=-\log \left(3.51 \times 10^{-4}\right)=3.455

BALLPARK CHECK

Because the pH is about 2, [ H3O+H_3O^+ ] and [ FF^- ] are about 10210^{-2} M and [HF] is about 0.25 M (0.250 M – 10210^{-2} M). The value of Ka is therefore about (10210^{-2} )(10210^{-2} )/0.25, or 4 × 10410^{-4} , and the pKaK_a is between 3 and 4. The ballpark check and the solution agree.

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