Calculate the pH of 0.0035 M Ba(OH)_2 solut | vedclass

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Question

(A) $Ba(OH)_2$ is a strong base and dissociates completely as: $Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^-$. The concentration of $OH^-$ ions is $[OH^-] = 2

Vedclass · Accepted Answer

$11.84$

Vedclass · Answer

(A) $Ba(OH)_2$ is a strong base and dissociates completely as: $Ba(OH)_2 ightarrow Ba^{2+} + 2OH^-$. The concentration of $OH^-$ ions is $[OH^-] = 2 	imes [Ba(OH)_2] = 2 	imes 0.0035 \ M = 0.007 \ M$. Now,$pOH = -\log[OH^-] = -\log(0.007) = -\log(7 	imes 10^{-3}) = 3 - \log(7) \approx 3 - 0.845 = 2.155$. Since $pH + pOH = 14$,we have $pH = 14 - 2.155 = 11.845 \approx 11.84$.

Calculate the $pH$ of $0.0035 \ M \ Ba(OH)_2$ solution.

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