# Question 3c224

Feb 3, 2015

The answer is $\text{23.0 mL}$ of disodium phosphate solution is needed for this particular buffer.

So, you know that the total volume of the buffer will be $\text{40.0 mL}$, or $\text{0.04 L}$. This means that the sum of the volumes of the two solutions must be

${V}_{1} + {V}_{2} = 0.04$ (1)

SInce you're dealing with a buffer, use the Henderson-Hasselbalch equation to determine another relationship between ${V}_{1}$ and ${V}_{2}$

$\left[p H\right] \left(h \texttt{p} : / s o c r a t i c . \mathmr{and} \frac{g}{c} h e m i s t r \frac{y}{a} c i \mathrm{ds} - \mathmr{and} - b a s e \frac{s}{t} h e - p h - c o n c e p t\right) = p K a + \log \left(\frac{\left[N {a}_{2} H P {O}_{4}\right]}{\left[N a {H}_{2} P {O}_{4}\right]}\right)$

$6.95 = 6.82 + \log \left(\frac{\left[N {a}_{2} H P {O}_{4}\right]}{\left[N a {H}_{2} P {O}_{4}\right]}\right) \implies \frac{\left[N {a}_{2} H P {O}_{4}\right]}{\left[N a {H}_{2} P {O}_{4}\right]} = 1.35$ (2)

Now, use the initial concentrations of the two solutions to find the relationship you need. For sodium phosphate, the buffer concentration can be written as

$\left[N a {H}_{2} P {O}_{4}\right] = {n}_{\text{sodium}} / \left({V}_{1} + {V}_{2}\right)$

The number of moles of sodium phosphate is

n_("sodium") = C * V_1 = "0.100 M" * V_1 = 0.100 * V_1

Likewise, the concentration of disodium phophate is

$\left[N {a}_{2} H P {O}_{4}\right] = {n}_{\text{disodium}} / \left({V}_{1} + {V}_{2}\right)$

Once again, the number of moles can be calculated using the initial concentration

n_("disodium") = C * V_2 = "0.100 M" * V_2 = 0.100 * V_2#

Plug all of this into equation (2)

$\frac{0.100 \cdot {V}_{2}}{{V}_{1} + {V}_{2}} \cdot \frac{{V}_{1} + {V}_{2}}{0.100 \cdot {V}_{1}} = 1.35$, or

${V}_{2} / {V}_{1} = 1.35 \implies {V}_{2} = 1.35 \cdot {V}_{1}$. Plug this into equation (1)

${V}_{1} + 1.35 \cdot {V}_{1} = 0.04 \implies {V}_{1} = \frac{0.04}{2.35} = 0.017$, which means that

${V}_{2} = 0.04 - 0.017 = 0.023$

Therefore, the volume of disodium phosphate $\text{0.100 M}$ solution you need for this particular buffer is

${V}_{2} = \text{0.023 L"= "23.0 mL}$