What is the pH of 10^{-4} N Ba(OH)_2 solutio | vedclass

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(B) $Ba(OH)_2$ is a strong base that dissociates as: $Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^-$. For a strong base,the normality $(N)$ is related to the m

Vedclass · Accepted Answer

$10$

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(B) $Ba(OH)_2$ is a strong base that dissociates as: $Ba(OH)_2 ightarrow Ba^{2+} + 2OH^-$. For a strong base,the normality $(N)$ is related to the molarity $(M)$ by the factor of acidity ($n$-factor). For $Ba(OH)_2$,the $n$-factor is $2$. Given normality $N = 10^{-4} \ N$. Since $N = M 	imes n	ext{-factor}$,we have $M = \frac{N}{2} = \frac{10^{-4}}{2} = 0.5 	imes 10^{-4} \ M$. The concentration of $OH^-$ ions is $[OH^-] = M 	imes n	ext{-factor} = 10^{-4} \ M$. $pOH = -\log[OH^-] = -\log(10^{-4}) = 4$. Using the relation $pH + pOH = 14$,we get $pH = 14 - 4 = 10$.

What is the $pH$ of $10^{-4} \ N$ $Ba(OH)_2$ solution?

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