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 27.0 ml of naoh. express your answer numerically.

Answer : The pH after 27.0 mL of NaOH added will be 2.05

Explanation : Given,

Concentration of acetic acid = 0.35 M

Volume of acetic acid = 52.0 mL = 0.052 L    (conversion used : 1 L = 1000 mL)

Concentration of NaOH = 0.40 M

Volume of NaOH = 27.0 mL = 0.027 L

First we have to calculate the moles of and .

The balanced chemical reaction is,

Initial moles         0.018         0.011               0

At eqm. moles   (0.018-0.011)  0                0.011

                            = 0.0007

Now we have to calculate the hydrogen ion concentration.

Now we have to calculate the pH.

Therefore, the pH after 27.0 mL of NaOH added will be 2.05