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?

  • $\begingroup$ You can use substitition, [B] = 0.5 - [A]. $\endgroup$
    – LDC3
    Mar 26, 2014 at 13:11
  • $\begingroup$ One thing has to be explained by the one who put up this question before a tangible answer without assumption can be made. The question mentions 100mL of c = 0.50M. What substance is c representing? It is molarity, but molarity of what chemical? Buffers have two chemicals in them, the weak acid and the salt which has the same anion (negatively charged species) as the weak acid. That anion is called the common ion. So if you say molarity is 0.50M, which of the two species are you referring to, the weak acid (which from your pKa should be H2PO4-) or the salt (which in this case should be HPO4--) $\endgroup$
    – user5558
    May 19, 2014 at 15:44

1 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.


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