$$3g$$ of acetic acid is added to $$250$$ mL of $$0.1$$ M HCl and the solution made up to $$500$$ mL. To $$20$$ mL of this solution $$12$$ mL of $$5M$$ NaOH is added. The pH of this solution is [ Given pKa of acetic acid $$=$$ $$4.74$$, molar mass of acetic acid $$=$$ $$60$$ $$g/mol$$, log $$3$$ $$=$$ $$0.4771$$]Neglect any changes in volume.

Question

$$3g$$ of acetic acid is added to $$250$$ mL of  $$0.1$$ M HCl and the solution made up to $$500$$ mL. To $$20$$ mL of this solution $$12$$ mL of $$5M$$ NaOH is added. The pH of this solution is [ Given pKa of acetic acid $$=$$ $$4.74$$, molar mass of acetic acid $$=$$ $$60$$ $$g/mol$$, log $$3$$ $$=$$ $$0.4771$$]Neglect any changes in volume.

Infinity Learn · Accepted Answer

m mole of acidic acid in $$20$$ mL $$=$$ $$2$$          M mole of HCl in $$20$$ mL $$=$$ $$1$$          M mole of NaOH $$=$$ $$2.5CH3COOH+NaOHremaining$$→$$CH3COONa+water$$     $$2$$                              $$3/2$$                          $$0$$                          $$0$$     $$0.5$$                             $$0$$                          $$3/2$$                         −$$pH=PKa+log3/22$$ $$=4.74+log3=4.74+0.48=5.22$$

3g of acetic acid is added to 250 mL of 0.1 M HCl and the solution made up to 500 mL. To 20 mL of this solution 12 mL of 5M NaOH is added. The pH of this solution is [ Given pKa of acetic acid = 4.74, molar mass of acetic acid = 60 g/mol, log 3 = 0.4771]Neglect any changes in volume.

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