pH of a 0.1 M monobasic acid is found to be | vedclass

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(B) Given,$pH = 2$,so $[H^+] = 10^{-2} = 0.01 \ M$.For a monobasic acid $HA \rightleftharpoons H^+ + A^-$,the degree of dissociation $\alpha$ is given

Vedclass · Accepted Answer

$0.11$

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(B) Given,$pH = 2$,so $[H^+] = 10^{-2} = 0.01 \ M$.For a monobasic acid $HA ightleftharpoons H^+ + A^-$,the degree of dissociation $\alpha$ is given by $[H^+] = C \alpha$.$0.01 = 0.1 	imes \alpha \implies \alpha = 0.1$.The van't Hoff factor $i = 1 + (n-1)\alpha$. Since $n=2$ (for $HA ightarrow H^+ + A^-$),$i = 1 + (2-1)(0.1) = 1.1$.The osmotic pressure $\pi = iCRT$.$\pi = 1.1 	imes 0.1 	imes RT = 0.11 \ RT$.

$pH$ of a $0.1 \ M$ monobasic acid is found to be $2$. Hence its osmotic pressure at a given temperature $T \ K$ is (in $RT$)

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