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(C) At the equivalence point,the salt sodium formate $(HCOONa)$ is formed. It is a salt of a weak acid and a strong base.The $pH$ of such a salt solut

Vedclass · Accepted Answer

$8.22$

Vedclass · Answer

(C) At the equivalence point,the salt sodium formate $(HCOONa)$ is formed. It is a salt of a weak acid and a strong base.The $pH$ of such a salt solution is given by the formula: $pH = 7 + 0.5 \ pK_a + 0.5 \ \log \ C$.First,calculate the volume of $NaOH$ required: $M_1V_1 = M_2V_2 \implies 0.1 \ M 	imes 25 \ mL = 0.1 \ M 	imes V_2 \implies V_2 = 25 \ mL$.Total volume of the solution $= 25 \ mL + 25 \ mL = 50 \ mL$.The concentration of the salt $(C)$ is: $C = \frac{0.1 \ M 	imes 25 \ mL}{50 \ mL} = 0.05 \ M$.Now,substitute the values into the $pH$ formula:$pH = 7 + 0.5 	imes 3.74 + 0.5 \ \log \ (0.05)$$pH = 7 + 1.87 + 0.5 \ \log \ (5 	imes 10^{-2})$$pH = 8.87 + 0.5 	imes (\log \ 5 - 2)$Since $\log \ 5 = \log \ (10/2) = 1 - 0.30 = 0.70$,$pH = 8.87 + 0.5 	imes (0.70 - 2) = 8.87 + 0.5 	imes (-1.30) = 8.87 - 0.65 = 8.22$.

Calculate the $pH$ at the equivalence point in the titration of $25 \ mL$ of $0.1 \ M$ formic acid with a $0.1 \ M$ $NaOH$ solution (given that $pK_a$ of formic acid $= 3.74$). $[\log \ 2 = 0.30]$

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