Calculate the moles of acid and conjugate base needed

Question

Question

Using the following salts listed below, calculate the moles of acid and conjugate base needed to make $V = \pu{100 mL}$ of $c = \pu{0.50 M}$, $pH \in 6.5, 6.7, 6.8$ phosphate buffer solutions. ($\mathrm pK_\mathrm a(\ce{H2PO4}) = 6.64$)

Molas masses of salts:

1. $\ce{KH2PO4} = 136.09 ~\pu{g/mol}$
2. $\ce{NaH2PO\cdot H2O} = \pu{137.99 g/mol}$
3. $\ce{Na2HPO4 \cdot 7H2O} = \pu{268.09 g/mol}$
4. $\ce{K2HPO4} = \pu{174.2 g/mol}$

I tried using the Henderson-Hasselbalch equation but I am stuck at the point where I have found the base-acid ratio and I need to find the moles of the acid and the moles of the base.

$$ \mathrm pH = \mathrm pK_\mathrm a + \log\left(\frac{[\text{Conjugate base}]}{[\text{Acid}]}\right)\\ n = \pu{100 ml} \times \pu{0.5 M} =\pu{ 0.05 mol} $$

For pH 6.5: $$ 6.5 = 6.64 + \log\frac{[B]}{[A]}\\ \frac{[B]}{[A]} = 0.72 $$ Here I got the base-acid ratio as 0.72 but what should I do next? I tried looking up for solutions and a person suggests using $A + B = 0.05$ But how do I know what A and B are just by knowing the ratio is 0.72?

Answer 1

Since it is a buffer solution, the volume is constant ($V =\pu{0.1 L}$), hence $$ \frac{c(b)}{c(a)} = \frac{n(b)}{n(a)}\overset{calc.}{=}0.72 $$ and with $$ n_{\text{tot}}=n(b)+n(a) \overset{\text{def}.}{=}\pu{0.05 mol}\\ n(a)=\frac1{1.72}\cdot\pu{0.05 mol}\\ n(b)=\frac{0.72}{1.72}\cdot\pu{0.05 mol} $$ Then $$m=M\cdot n$$ and you should be fine.