When 50 ml of 0.1 M $\mathrm{NaOH}$ is added to 5 ml of 0.05 M ${\mathrm{CH}}_3\mathrm{COOH}$ solution the pH of the solution is

0.1M of NaOH is present in 50ml solution.

0.05M of CH3​COOH is present in 50ml solution.

CH3​COOH - weak acid and NaOH - strong base

The reaction between CH₃​COOH and NaOH can be written as,

CH3​COOH +Na⁺OH⁻→CH₃​COO⁻Na⁺+H₂​O

Since we know that strong bases and salts are easily dissociated as ions.

It is known that the product of volume in milliliters and molarity gives the number of millimoles of the acid or base.

The number of millimoles of acetic acid present in the solution=50×0.05

                                                                                                         =2.5

The number of millimoles of sodium hydroxide present in the solution=50×0.1

                                                                                                                    =5

Thus, 2.5m.mol of NaOH neutralises 2.5m.mol of CH₃​COOH forming CH₃​COONa (sodium acetate) salt.

Therefore, the number of millimoles present in sodium acetate salt= 2.5m.mol

This leads to the formation of the weak acid and its salt buffer since acetic acid is not completely neutralized by sodium hydroxide and the remaining 2.5m.mol of excess sodium hydroxide is still present.

Molarity of excess sodium hydroxide present in 100ml solution (50ml NaOH+50ml CH3​COOH )=$\frac{25}{100}$=0.025M

       It is known that,

pOH= −log[OH⁻]

⇒pOH=−log[0.025]

⇒p^OH=1.6021

It is also known that p^H+p^OH=14

⇒pH=14−pOH

⇒p^H =14–1.6021

⇒p^H =12.3979