At 30,^{circ}C temperature,what is the ratio | vedclass

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(A) The dissociation constant of water $(K_a)$ is related to the ionic product of water $(K_w)$ by the equation: $K_a = \frac{K_w}{[H_2O]}$.At $30\,^{

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$18:1000$

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(A) The dissociation constant of water $(K_a)$ is related to the ionic product of water $(K_w)$ by the equation: $K_a = \frac{K_w}{[H_2O]}$.At $30\,^{\circ}C$,the molar concentration of water $[H_2O]$ is approximately $55.5 \, M$ (calculated as $\frac{1000 \, g/L}{18 \, g/mol} \approx 55.5 \, mol/L$).Therefore,the ratio $\frac{K_a}{K_w} = \frac{1}{[H_2O]} = \frac{1}{55.5}$.Converting this to a ratio of $18:1000$: $\frac{1}{55.5} \approx 0.018 = \frac{18}{1000}$.

At $30\,^{\circ}C$ temperature,what is the ratio of the dissociation constant of water $(K_a)$ to the ionic product of water $(K_w)$?

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