# Question 15.9: The pH of 0.250 M HF is 2.036. What are the values of Ka and......

The pH of 0.250 M HF is 2.036. What are the values of $K_{a}$ and $pK_{a}$ for hydrofluoric acid?

STRATEGY

Use the general procedure for solving equilibrium problems described in Section 14.5. First, write the balanced equation for the dissociation equilibrium. Then, add known quantities to the “initial,” “change,” and “equilibrium” rows in the equilibrium table. Define x as the concentration of HF that dissociates. Because x equals the $H_{3}O^{+}$ concentration, its value can be calculated from the pH. Finally, substitute the equilibrium concentrations into the equilibrium equation to obtain the value of $K_{a}$ and take the negative log of $K_{a}$ to obtain the $pK_{a}$.

 IDENTIFY Known Unknown Concentration of HF (0.250 M) $K_{a}, pK_{a}$ pH = 2.036
Step-by-Step
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We can calculate the value of x from the pH:

$[H_{3}O^{+}]=antilog (-pH)=10^{-pH}=10^{-2.036}=9.20\times 10^{-3} M$

The other equilibrium concentrations are

$[F^{-}]=x=9.20\times 10^{-3} M$

[HF] = 0.250 – x = 0.250 – 0.009 20 = 0.241 M

Substituting these concentrations into the equilibrium equation gives the value of $K_{a}$:

$K_{a}=\frac{[H_{3}O^{+}][F^{-}]}{[HF]}=\frac{(x)(x)}{(0.250 – x)}=\frac{(9.20\times 10^{-3})(9.20\times 10^{-3})}{0.241}=3.51\times 10^{-4}$

$pK_{a}=- log K_{a}=-log (3.51\times 10^{-4})=3.455$
 HF(aq) + $H_{2}O(l)$ $\xrightleftharpoons{}$ $H_{3}O^{+}(aq)$ + $F^{-}(aq)$ Initial concentration (M) 0.250 ∼0* 0 Change (M) -x +x +x Equilibrium concentration (M) (0.250 – x) x x *A very small concentration of $H_{3}O^{+}$ is present initially because of the dissociation of water.

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