Answer to Question #155415 in General Chemistry for C

Question

If 20.0 mL of a 0.30 M acetic acid (CH3COOH) is titrated with 10.0 mL of 0.30 M NaOH solution. 

     Calculate the pH of the resulting solution. The Ka of acetic acid is 1.8 x 10-5.

Answer 1

CH3COOH+NaOH=CH3COONa+H2O

n=C×V

n(CH3COOH)=0.30×0.02=0.006(mol)

n(NaOH)=0.30×0.01=0.003(mol)

So, after reaction

0.006-0.003=0.003(mol) of CH3COOH are left.

At the same time 0.003 mol of CH3COONa is formed. The mixture of a weak acid and it's salt is a buffer solution, and pH should be found from Henderson-Hasselbalch equation:

pH=pKa+logC(CH3COONa)/C(CH3COOH).

pKa=-log1.8×10^-5=4.74

C=n/V

V=0.01+0.02=0.03(L)

C(CH3COOH)=C(CH3COONa)=0.003/0.03=0.1(mol/L)

pH=4.74+log0.1/0.1=4.74+log1=4.74+0=4.74.