# A 250.0 mL buffer solution is 0.250 M in acetic acid and 0.250 M in sodium acetate. What is the initial pH of this solution? What is the pH after addition of 0.0050 mol of HCl? What is the pH after addition of 0.0050 mol of NaOH?

Apr 5, 2016

$\left(a\right) \text{ } 4.74$

$\left(b\right) \text{ } 4.39$

$\left(c\right) \text{ } 5.09$

#### Explanation:

$\left(a\right)$

Ethanoic acid is a weak acid and dissociates:

$C {H}_{3} C O O {H}_{\left(a q\right)} r i g h t \le f t h a r p \infty n s C {H}_{3} C O {O}_{\left(a q\right)}^{-} + {H}_{\left(a q\right)}^{+}$

For which:

${K}_{a} = \frac{\left[C {H}_{3} C O {O}_{\left(a q\right)}^{-}\right] \left[{H}_{\left(a q\right)}^{+}\right]}{\left[C {H}_{3} C O O {H}_{\left(a q\right)}\right]} = 1.8 \times {10}^{- 5} \text{mol/l}$ at ${25}^{\circ} \text{C}$

Rearranging gives:

$\left[{H}_{\left(a q\right)}^{+}\right] = {K}_{a} \times \frac{\left[C {H}_{3} C O O {H}_{\left(a q\right)}\right]}{\left[C {H}_{3} C O {O}_{\left(a q\right)}^{-}\right]}$

I am going to assume that equilibrium concentrations have been given so:

$\left[{H}_{\left(a q\right)}^{+}\right] = 1.8 \times {10}^{- 5} \times \frac{\cancel{0.25}}{\cancel{0.25}}$

$p H = - \log \left[{H}_{\left(a q\right)}^{+}\right] = - \log \left(1.8 \times {10}^{- 5}\right)$

$\textcolor{red}{p H = 4.74}$

$\left(b\right)$

The buffer contains a large reserve of the co - base which can absorb the addition of a small amount of ${H}^{+}$ ions:

$C {H}_{3} C O {O}_{\left(a q\right)}^{-} + {H}_{\left(a q\right)}^{+} \rightarrow C {H}_{3} C O O {H}_{\left(a q\right)}$

The initial no. of moles of $C {H}_{3} C O {O}_{\left(a q\right)}^{-}$ is given by:

$n = c \times v = 0.25 \times \frac{250}{1000} = 0.0625$

The no. of moles of ${H}^{+}$ added we are told, is $0.005$.

From the equation we can see that they react in a 1:1 mole ratio.

So the no. moles $C {H}_{3} C O {O}_{\left(a q\right)}^{-}$ remaining is given by:

${n}_{C {H}_{3} C O {O}^{-}} = 0.0625 - 0.005 = 0.0575$

The no. moles of $C {H}_{3} C O O H$ will have increased by the same amount.

The initial no. moles of $C {H}_{3} C O O H$ is given by:

$n = c \times v = 0.25 \times \frac{250}{1000} = 0.0625$

So the total moles $C {H}_{3} C O O {H}_{\left(a q\right)}$ remaining is given by:

${n}_{C {H}_{3} C O O H} = 0.0625 + 0.005 = 0.13$

We don't know the new total volume but since this is common to acid and co - base it will cancel so we can write:

$\left[{H}_{\left(a q\right)}^{+}\right] = 1.8 \times {10}^{- 5} \times \frac{0.13}{0.0575} = 4.069 \times {10}^{- 5}$

$p H = - \log \left(4.069 \times {10}^{- 5}\right)$

$\textcolor{red}{p H = 4.39}$

$\left(c\right)$

If a small amount of $O {H}_{\left(a q\right)}^{-}$ ions is added these will react with $C {H}_{3} C O O {H}_{\left(a q\right)}$:

$C {H}_{3} C O O {H}_{\left(a q\right)} + O {H}_{\left(a q\right)}^{-} \rightarrow C {H}_{3} C O {O}_{\left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)}$

We have already calculated the initial moles of $C {H}_{3} C O O H$ to be $0.0625$.

You can see from the equation that they react with $O {H}^{-}$ in a 1:1 molar ratio so the no. moles remaining is given by:

${n}_{C {H}_{3} C O O H} = 0.0625 - 0.005 = 0.0575$

The no. moles of $C {H}_{3} C O {O}^{-}$ will have increased by the same amount.

$\therefore {n}_{C {H}_{3} C O {O}^{-}} = 0.0625 + 0.005 = 0.13$

Again, since the new total volume is constant we can write:

$\left[{H}_{\left(a q\right)}^{+}\right] = 1.8 \times {10}^{- 5} \times \frac{0.0575}{0.13}$

$\left[{H}_{\left(a q\right)}^{+}\right] = 0.796 \times {10}^{- 5}$

$\textcolor{red}{p H = 5.09}$