Question

You have 500 mL of a buffer solution containing 0.3 M Acetic acid and 0.2 M acetate . The PH of the buffer is 4.57. What will be |Delta PH| of this solution after the addition of 20 mL of 1 M KOH? Pka of the weak acid is 4.75

Answer 1

The additional volume of KOH will react with acetic acid in the buffer according the following equation:

CH₃COOH + KOH = CH₃COOK + H₂O

We have to calculate the amount of added KOH in moles:

$n(KOH)=V(KOH)*C_{KOH}=0.02L*1mol/L=0.02mol$

We also have to calculate the amounts in moles of CH₃COOH and acetate anion in the buffer:

$n(acetate)=C_{acetate}*V_{buffer}=0.2mol/L*0.5L=0.1mol$

$n(acid)=C_{acid}*V_{buffer}=0.3mol/L*0.5L=0.15mol$

One can easily note, that some amount of acid will be consumed, while that of acetate anion will be increase:

$n(acetate)_{new}=n(acetate)+n(KOH)=0.10mol+0.02mol=0.12mol$

$n(acid)_{new}=n(acid)-n(KOH)=0.15mol-0.02mol=0.13mol$

In the first approximation, the total volume of the new buffer will be the additive of the old volume and the added one:

$V(buffer)_{new}=V(buffer)+V(KOH)=0.500L+0.020L=0.520L$

Let us now calculate the novel concentrations of acid and acetate anion:

$C(acid)_{new}=\frac{n(acid)_{new}}{V(buffer)_{new}}=\frac{0.13mol}{0.520L}=0.25M$

$C(acetate)_{new}=\frac{n(acetate)_{new}}{V(buffer)_{new}}=\frac{0.12mol}{0.520L}=0.23M$

Now it is right time to evaluate the new pH value:

$pH_{new}=pK_{a}+lg(\frac{C(acetate)_{new}}{C(acid)_{new}})=4.75+lg(\frac{0.23M}{0.25M})=4.71$

Thus, the absolute value of the pH delta is:

|$\Delta$ pH| = 4.71 - 4.57 = 0.14