Question

How much 0.541 M KOH will be needed to raise the pH of 0.391 L of...

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?

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 CO32-.

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

pH= pKa2 + log([CO32-]/[HCO3-]) putting values into it we get

9.498= 10.319 + log([CO32-]/[HCO3-]), solving it gives ([CO32-]/[HCO3-]) = 10-0.821 =0.151 -(i)

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

= moles of CO32- + moles of HCO3- =moles of CO32-(1+1/0.151) (using (i) above), solving it gives moles of CO32- = 0.237 moles

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

Moles of KOH required = moles of HCO3- + 2× moles of CO32- =1.573+2×0.237 = 2.047 moles

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

Comment in case of any doubt.