# A 52.0 mL volume of 0.35 M CH3COOH (Ka = 1.8*10^-5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of...

A 52.0 mL volume of 0.35 M CH3COOH (Ka = 1.8*10^-5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of NaOH.

Initial moles of CH3COOH present in 52.0 mL of 0.35 M is :
52.0 mL * ( 1 L / 1000 mL ) * 0.35 M = 0.018 mol
Initial  moles of NaOH present in 23.0 mL of 0.40 M is :
23.0 mL * ( 1 L / 1000 mL ) * 0.40 M = 9.2*10^-3 mol
CH3COOH   + NaOH -------> CH3COONa   + H2O
I(mol)   0.018           9.2*10-3                              0                 -
C(mol  - 9.2*10-3    -9.2*10-3                      +9.2*10-3
-----------------------------------------------------------
E(mol) 8.8*10-3                0                          9.2*10-3
CH3COOH   -------> CH3COO-   + H+
Moles:     8.8*10-3                       9.2*10-3
Ka  = [CH3COO- ] [ H+] / [CH3COOH ]
[ H+] = Ka  [CH3COOH ] / [CH3COO- ]
= (1.8*10-5 ) (8.8*10-3  ) / (9.2*10-3  )
= 1.7*10-5
pH = -log[H+] = 4.77
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