A52.0-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.

 52.0ml of 0.35M CH3COOH : 0.052 L(0.35M) = .0182 mol of CH3COOH. 

31.0ml of 0.40M NaOH : .031 L(0.40M) = .0124 mol of NaOH. 

After the  reaction, .0124 Mol CH3COO- is generated and .058 mol CH3COOH is left un-reacted. The concentration would be 12.4/V and 5.8/V, respectively. Therefore:

 
pH = -log([H+]) = -log(Ka*[CH3COOH]/[CH3COO-]) 
= -log(1.8x10^-5*5.8/12.4) = 5.07