Question e6995

Feb 4, 2015

You'd need $\text{41.7 mL}$ of citric acid for this particular buffer.

You need to find two equations that you can use to determine the volume of the citric acid solution. The first one will be

V_("buffer") = V_1 + V_2 = "125 mL" = "0.125 L" (1)

The total volume of the buffer solution will be equal to the sum of the two solutions mixed together - ${V}_{1}$ is the volume of the citric acid solution, while ${V}_{2}$ is the volume of the sodium citrate solution.

Next, use the Henderson-Hasselbalch equation

$p {H}_{\text{solution}} = p K a + \log \left(\frac{\left[{C}_{6} {H}_{5} {O}_{7}^{3 -}\right]}{\left[{C}_{6} {H}_{8} {O}_{7}\right]}\right)$

$3.45 = 3.15 + \log \left(\frac{\left[{C}_{6} {H}_{5} {O}_{7}^{3 -}\right]}{\left[{C}_{6} {H}_{8} {O}_{7}\right]}\right) \implies \frac{\left[{C}_{6} {H}_{5} {O}_{7}^{3 -}\right]}{\left[{C}_{6} {H}_{8} {O}_{7}\right]} = 2.0$ (2)

Now, the concentration of the citric acid in the buffer is equal to

${C}_{\text{citric") = n_("citric}} / \left({V}_{1} + {V}_{2}\right)$

The number of moles of citric acid can be determined from the initial concentration

n_("citric") = C * V_1 = "0.150 M" * V_1 = 0.150 * V_1

Likewise, the concentration of the citrate is

${C}_{\text{citrate") = n_("citrate}} / \left({V}_{1} + {V}_{2}\right)$, and

n_("citrate") = C * V_2 = "0.150 M" * V_2 = 0.150 * V_2#

Plug all of this into equation (2) and you'll get

$\frac{0.150 \cdot {V}_{2}}{{V}_{1} + {V}_{2}} \cdot \frac{{V}_{1} + {V}_{2}}{0.150 \cdot {V}_{1}} = 2.0$, or

${V}_{2} / {V}_{1} = 2 \implies {V}_{2} = 2 \cdot {V}_{1}$. Plug this into equation (1)

${V}_{1} + 2 \cdot {V}_{1} = 0.125 \implies {V}_{1} = \frac{0.125}{3} = 0.0417$, which means that

${V}_{2} = 0.125 - 0.0417 = 0.0833$

Therefore, the volume for the citric acid solution will need to be

${V}_{1} = \text{0.0417 L" = "41.7 mL}$