pH of a buffer solution calculation
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:

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

$\mathrm{pH}=3.75$