# A 1.00 L buffer solution is 0.150 M in HC_7H_5O_2 and 0.250 M in LiC_7H_5O_2. How do you calculate the pH of the solution after the addition of 100.0 mL of 1.00 M HCl? The Ka for HC_7H_5O_2 is 6.5 × 10^–5.

Apr 11, 2016

You can do it like this:

#### Explanation:

It looks like the acid is benzoic acid ${C}_{6} {H}_{5} C O O H$

For short - hand I'll call this $H A$

Benzoic acid dissociates:

$H {A}_{\left(a q\right)} r i g h t \le f t h a r p \infty n s {H}_{\left(a q\right)}^{+} + {A}_{\left(a q\right)}^{-} \text{ } \textcolor{red}{\left(1\right)}$

The lithium salt provides a large reserve of the co - base ${A}^{-}$ by dissociating completely:

$L i {A}_{\left(s\right)} \rightarrow L {i}_{\left(a q\right)}^{+} + {A}_{\left(a q\right)}^{-} \text{ } \textcolor{red}{\left(2\right)}$

The initial no. moles of ${A}^{-}$ is given by:

$n {A}_{\text{init}}^{-} = 1 \times 0.25 = 0.25$

The no. ${H}^{+}$ added is given by:

$n {H}^{+} = 1 \times \frac{100}{1000} = 0.1$

From $\textcolor{red}{\left(1\right)}$ you can see that these ${H}^{+}$ ions will be absorbed by the large reserve of ${A}^{-}$ ions shifting the equilibrium to the left thus "buffering" the pH.

You can see that they react in a 1:1 molar ratio so that the no. of moles of ${A}^{-}$ remaining is given by:

$n {A}^{-} = 0.25 - 0.1 = 0.15$

Each mole of ${H}^{+}$ added will form 1 mole of $H A$ so no. moles $H A$ formed $= 0.1$

The initial moles of $H A$ is given by:

$n H {A}_{\text{init}} = 0.15 \times 1 = 0.15$

So the total moles $H A$ is given by:

$n H A = 0.1 + 0.15 = 0.25$

The expression for the acid dissociation constant ${K}_{a}$ is given by:

${K}_{a} = \frac{\left[{H}_{\left(a q\right)}^{+}\right] \left[{A}_{\left(a q\right)}^{-}\right]}{\left[H {A}_{\left(a q\right)}\right]}$

Rearranging:

$\left[{H}_{\left(a q\right)}^{+}\right] = {K}_{a} \times \frac{\left[H {A}_{\left(a q\right)}\right]}{\left[{A}_{\left(a q\right)}^{-}\right]}$

I will ignore any changes in concentration due to the dissociation of the acid as it is very small.

The total volume is common to both acid and co - base and will cancel so we can write:

$\left[{H}_{\left(a q\right)}^{+}\right] = 6.5 \times {10}^{- 5} \times \frac{0.25}{0.15} = 1.083 \times {10}^{- 4}$

$p H = - \log \left(1.083 \times {10}^{- 4}\right)$

$\textcolor{red}{p H = 3.96}$