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

for points guys

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explanation:

smoke is the matter of a gas