pH of 10^{-8} M Ba(OH)_2 solution will be | vedclass

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(C) $Ba(OH)_2$ is a strong base and dissociates as: $Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^-$. Since the concentration of $Ba(OH)_2$ is $10^{-8} \ M$,the

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$7.08$

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(C) $Ba(OH)_2$ is a strong base and dissociates as: $Ba(OH)_2 ightarrow Ba^{2+} + 2OH^-$. Since the concentration of $Ba(OH)_2$ is $10^{-8} \ M$,the concentration of $OH^-$ ions from the base is $[OH^-]_{base} = 2 	imes 10^{-8} \ M$. At such low concentrations,the contribution of $OH^-$ from water $(10^{-7} \ M)$ cannot be ignored. Total $[OH^-] = [OH^-]_{base} + [OH^-]_{water} = 2 	imes 10^{-8} + 10^{-7} = 1.2 	imes 10^{-7} \ M$. Now,$pOH = -\log[OH^-] = -\log(1.2 	imes 10^{-7}) = 7 - \log(1.2) \approx 7 - 0.079 = 6.921$. Since $pH + pOH = 14$,$pH = 14 - 6.921 = 7.079 \approx 7.08$.

$pH$ of $10^{-8} \ M$ $Ba(OH)_2$ solution will be

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