If the pH of 0.10 M monobasic acid at 298 K | vedclass

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(C) Given: $pH$ for monobasic acid $= 5.0$The dissociation reaction for a monobasic acid $(HA)$ is:$HA(aq) \rightleftharpoons H^+(aq) + A^-(aq)$Initia

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$9$

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(C) Given: $pH$ for monobasic acid $= 5.0$The dissociation reaction for a monobasic acid $(HA)$ is:$HA(aq) \rightleftharpoons H^+(aq) + A^-(aq)$Initial concentration: $0.10 \ M$,$0$,$0$Equilibrium concentration: $(0.10 - x)$,$x$,$x$Since $pH = -\log[H^+] = 5.0$,we have:$[H^+] = 10^{-5} \ M$Therefore,$x = 10^{-5} \ M$.The acid dissociation constant $K_a$ is given by:$K_a = \frac{[H^+][A^-]}{[HA]} = \frac{x^2}{0.10 - x}$Since $x = 10^{-5}$ is very small compared to $0.10$,we can approximate $0.10 - x \approx 0.10$:$K_a = \frac{(10^{-5})^2}{0.10} = \frac{10^{-10}}{10^{-1}} = 10^{-9}$Now,$pK_a = -\log(K_a) = -\log(10^{-9}) = 9$.

If the $pH$ of $0.10 \ M$ monobasic acid at $298 \ K$ is $5.0$,the value of $p K_a$ at the same temperature is

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