# Question dd6ba

Sep 23, 2015

Here's how I would do it.

#### Explanation:

You already know what your buffer must contain and what the total volume of the solution must be, so all you really need to do is work backward to find the procedure needed to make this buffer.

So, the total volume of the buffer is $\text{900 mL}$. This means that you can use the target molarities of the monosodium phosphate, ${\text{NaH"_2"PO}}_{4}$, and disodium phosphate, ${\text{Na"_2"HPO}}_{4}$, to find how many moles of each you need to have present in the buffer.

$C = \frac{n}{V} \implies n = C \cdot V$

For monosodium phosphate you get

${n}_{1} = 87.7 \cdot {10}^{- 3} \text{M" * 900 * 10^(-3)"L" = "0.07893 moles}$

For disodium phosphate you get

${n}_{2} = 12.3 \cdot {10}^{- 3} \text{M" * 900 * 10^(-3)"L" = "0.01107 moles}$

Next, use the molar masses of the two salts to find how many grams of each would contain these number of moles.

The interesting thing here is that the molecular weight you provided for the monosodium phosphate is actually that of monosodium phosphate monohydrate, $\text{NaH"_2"PO"_4 * "H"_2"O}$.

Since I don't know if this is a mistake or not, I'll assume that you're dealing with the hydrate and use the molecular weight you provided.

So, for monosodium phosphate monohydrate (?) you get

0.07893color(red)(cancel(color(black)("moles"))) * "137.99 g"/(1color(red)(cancel(color(black)("mole")))) = "10.89 g"

and for disodoium phosphate (anhydrous this time) you get

0.01107color(red)(cancel(color(black)("moles"))) * "141.96 g"/(1color(red)(cancel(color(black)("mole")))) = "1.571 g"

Now for the thimerasol. You know that the buffer must contain $\text{0.02%}$, I assume by mass, thimerasol.

If you take the density of the buffer to be equal to that of pure water at ${20}^{\circ} \text{C}$, you can say that the buffer will have a total mass of

http://antoine.frostburg.edu/chem/senese/javascript/water-density.html

900color(red)(cancel(color(black)("mL"))) * "0.9982071 g"/(1color(red)(cancel(color(black)("mL")))) = "898.4 g"#

This means that the mass of thimerasol that you need to dissolve in the buffer is

$\text{%m/m" = m_"thimerasol"/m_"sol} \times 100$

${m}_{\text{thimerasol" = (m_"sol" xx "%m/m")/100 = ("898.4 g" * 0.02)/100 = "0.180 g}}$

So, you're good to go. Start by weighing $\text{10.89 g}$ of solid monosodium phosphate monohydrate (?). Dissolve it in $\text{200 mL}$ of water.

Next, weigh $\text{1.571 g}$ of solid disodium phosphate and dissolve it in $\text{200 mL}$ of water.

Add these two solutions together and dissolve $\text{0.180 g}$ of thimerasol in the resulting solution.

Finally, add enough water until the total volume of solution is $\text{900 mL}$.

I didn't take into account significant figures here, especially since I'm sure that the volume of the solution will not have one sig fig.