Question

How much 0.541 M KOH will be needed to raise the pH of 0.391 L of 4.63 M carbonic acid (H₂CO₃) to a pH of 9.498?

Answer 1

For carbonic acid pKa1 =6.377 and pKa2= 10.319.

Given pH value is very close to pKa2 as compared to pKa1 value. So at this pH H2CO3 form of carbonic acid will be negligible as compared to its other forms and the buffer will, thus, be of HCO₃^- and CO₃^2-.

Now using Henderson Hasselbalch equation for this buffer at pH =9.498, we have:

pH= pKa₂ + log([CO₃^2-]/[HCO₃^-]) putting values into it we get

9.498= 10.319 + log([CO₃^2-]/[HCO₃^-]), solving it gives ([CO₃^2-]/[HCO₃^-]) = 10^-0.821 =0.151 -(i)

Initially moles of H2CO3 = 0.391L×4.63M = 1.81 mol

= moles of CO₃^2- + moles of HCO3^- =moles of CO₃^2-(1+1/0.151) (using (i) above), solving it gives moles of CO₃^2- = 0.237 moles

Moles of HCO₃^- =1.81-0.237 =1.573 moles.

Moles of KOH required = moles of HCO₃^- + 2× moles of CO₃^2- =1.573+2×0.237 = 2.047 moles

Now V_NaOH×0.541 M = 2.047 moles, solving it gives V_NaOH = 3.783 litres.

Comment in case of any doubt.