What is the pH of a 0.005 M Ba(OH)_2 solutio | vedclass

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Question

(A) $Ba(OH)_2$ is a strong base and dissociates completely as: $Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^-$. Since the concentration of $Ba(OH)_2$ is $0.005

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) $Ba(OH)_2$ is a strong base and dissociates completely as: $Ba(OH)_2 ightarrow Ba^{2+} + 2OH^-$. Since the concentration of $Ba(OH)_2$ is $0.005 \ M$,the concentration of $OH^-$ ions is $[OH^-] = 2 	imes 0.005 \ M = 0.01 \ M = 10^{-2} \ M$. Now,calculate $pOH$: $pOH = -\log[OH^-] = -\log(10^{-2}) = 2$. Finally,calculate $pH$ using the relation: $pH + pOH = 14$. $pH = 14 - 2 = 12$.

What is the $pH$ of a $0.005 \ M$ $Ba(OH)_2$ solution?

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