Question

What will the pH of the solution be when 0.4091 L of 1.1 M KOH are added to the 551 mL of 0.501 M carbonic acid?

Answer 1

2KOH + H2CO3 ------> K2CO3 + H2O

No of mole of KOH = (1.1mol/1000ml)×409.1ml = 0.45001

No of mole of H2CO3 in original solution = (0.501mol/1000ml)×551ml =0.27605

0.27605mol of H2CO3 react with 0.27605 mol of KOH to form 0.27605mol of KHCO3

remaining mol of KOH = 0.45001 - 0.27605 = 0.17396

0.17396mol of KOH react with 0.173968 mol of KHCO3 to form 0.17396 mol of K2CO3

No of mole of HCO3- remaining =0.10209

No of mole of CO32- formed = 0.17396

Total volume = 409.1ml + 551ml = 960.1ml

[ HCO3-] = (0.10209mol/960.1ml)×1000ml =0.1063M

[CO32-] = (0.17396mol/960.1ml)×1000ml = 0.1812M

pK2 of Carbonic acid = 10.33

pH = 10.33 + log(0.1812/0.1063)

= 10.33 + 0.23

= 10.56