Question

You are titrating 0.100 L of 0.100 M carbonic acid (shown below) with 0.500 M NaOH.

What will the pH be when 23.0 mL of NaOH have been added? pkas: 6.05 and 10.64

Answer 1

H2CO3----> H+ +HCO3-

HCO3- ----> H+ CO3-2

Pka1= 6.05,Ka1=8.91*10-7 Ka2= 10^(-10.64)=2.3*10-11

The reactino between H2CO3 and NaOH is H2CO3+2NaOH----> Na2CO3 +H2O

for every 1 mole of NaOH, we require 2 moles of NaOH for complete reaction

mole of H2CO3 initially= 0.1*0.1= 0.01 moles Moles of NaOH= 0.5*23/1000=0.0115

All the NaOH gets consumed and moles of H2CO3 remaining =0.01-0.0115/2=0.00425moles

Volume after mixing =0.1+23/1000 =0.123

Concentration of H2CO3= 0.00425/0.123=0.003455M

from Reactin H2CO3---> H+ HCO3-

Initailly [H2CO3] =0.003455 M

let x= drop in concentration of H2CO3 to reach equilibrium

At equilibrium   [H+] =[HCO3-] = x, [H2CO3] = 0.003455-x

Ka1= x2/(0.003455-x) =4.3*10^-7

since Ka1 is very large we can assumed 0.003455-x =0.003455

x2 =4.3*10-7*0.003455=1.49*10^-9

x= 3.854*10^-5

pH= -log(3.854*10^-5)=4.14

Since Ka1>>Ka2, we can neglect the H+ from the second