pH of a buffer solution calculation Calculate the pH of a mixture of $75\:\mathrm{cm^3}$ of methanoic acid solution of concentration $0.10\:\mathrm{mol\:dm^{-3}}$ and $75\:\mathrm{cm^3}$ of sodium hydroxide solution of concentration $0.050\:\mathrm{mol\:dm^{-3}}$. The $\mathrm{p}K_\mathrm{a}$ of methanoic acid is $3.75$.


I got $0.00375\:\mathrm{mol}$ of $\ce{HCOONa}$, the salt, made. The concentration of that is $\dfrac{0.00375}{150/1000}=0.025\:\mathrm{mol\:dm^{-3}}$(?)

Using Henderson–Hasselbalch: $\mathrm{pH}=3.75+\log\dfrac{0.025}{0.10}=3.15$

The answer says $3.75$, is that a typo?
One will need the Henderson Hasselbach equation which is given as

$$\text{pH}=pK_a+\log\frac{\ce{A-}}{\ce{HA}}$$

where HA and A- are the concentrations of the conjugate acid and conjugate base.

First we find our moles of each.

$\mathrm{{mol_ {acid}}=75~cm^3\times(0.10~mol~dm^{-3})\times(10cm)^{-3}\ dm^{3} = 0.0075\ mol\ \times \dfrac{1000\ mmol}{1\ mol} = \ 750~mmol}$

same thing for base: $\mathrm {mol_{base} = 375} $ mmol

Set up an ICA (not ICE) to handle your limiting reagent: (Write down the chemical equation if this is hard to see. One should react acid with the best base available)


$\ce{HA}$: $\pu{750 mmol -375 mmol} = \pu{375 mmol}$
$\ce{OH-}$: $\pu{375 mmol - 375 mmol} = \pu{0}$
$\ce{H2O}$: --------------------
$\ce{A-}$: $\pu{0 + 375 mmol} = \pu{375 mmol}$


In the Henderson Hasselbach equation moles can be used instead of molarity because both conjugates are in the same volume so they cancel:

$$\mathrm{pH}=3.75+\log\dfrac{375\ \mathrm{mmol\ \ce{A-}}}{375\ \mathrm{mmol\ } \ce{HA}}$$

$\mathrm{pH}=3.75$