The pH of an aqueous solution of H_2CO_3 is 3 | vedclass

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(A) For a weak diprotic acid like $H_2CO_3$,the first dissociation step is $H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$.The equilibrium constant is ${K_

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$5 	imes 10^{-4} \ M$

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(A) For a weak diprotic acid like $H_2CO_3$,the first dissociation step is $H_2CO_3 ightleftharpoons H^+ + HCO_3^-$.The equilibrium constant is ${K_{a_1}} = \frac{[H^+][HCO_3^-]}{[H_2CO_3]} = 10^{-3}$.Given $pH = 3.3$,we have $[H^+] = 10^{-3.3} \approx 5 	imes 10^{-4} \ M$.Since the dissociation of $H_2CO_3$ is the primary source of $H^+$ ions and $[H^+] \approx [HCO_3^-]$ (assuming negligible contribution from the second dissociation step due to very small ${K_{a_2}}$),the concentration $[HCO_3^-]$ is approximately equal to $[H^+]$.Therefore,$[HCO_3^-] \approx 5 	imes 10^{-4} \ M$.

The $pH$ of an aqueous solution of $H_2CO_3$ is $3.3$. If ${K_{a_1}} = {10^{-3}}$ and ${K_{a_2}} = {10^{-13}}$,then the concentration of $[HCO_3^-]$ is:

Similar Questions

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