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(C) The initial concentration of $Ba(OH)_2$ is $0.01 \, M$. Since $Ba(OH)_2$ dissociates as $Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^-$,the concentration o

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) The initial concentration of $Ba(OH)_2$ is $0.01 \, M$. Since $Ba(OH)_2$ dissociates as $Ba(OH)_2 ightarrow Ba^{2+} + 2OH^-$,the concentration of $OH^-$ ions is $2 	imes 0.01 \, M = 0.02 \, M$.Using the dilution formula $M_1V_1 = M_2V_2$ for the $OH^-$ concentration:$0.02 \, M 	imes 50 \, mL = M_2 	imes (50 \, mL + 50 \, mL)$$M_2 = \frac{0.02 	imes 50}{100} = 0.01 \, M$.Thus,$[OH^-] = 0.01 \, M = 10^{-2} \, M$.$pOH = -\log[OH^-] = -\log(10^{-2}) = 2$.Since $pH + pOH = 14$,we have $pH = 14 - 2 = 12$.

$50 \, mL$ water is added to a $50 \, mL$ solution of $Ba(OH)_2$ of strength $0.01 \, M$. The $pH$ value of the resulting solution will be

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