0.837 g of Ba(OH)_2 is dissolved in 100 mL | vedclass

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(A) $1$. Calculate the molar mass of $Ba(OH)_2$: $137 + 2 	imes (16 + 1) = 137 + 34 = 171 \ g/mol$. $2$. Calculate the number of moles of $Ba(OH)_2$:

Vedclass · Accepted Answer

$12.91$

Vedclass · Answer

(A) $1$. Calculate the molar mass of $Ba(OH)_2$: $137 + 2 	imes (16 + 1) = 137 + 34 = 171 \ g/mol$. $2$. Calculate the number of moles of $Ba(OH)_2$: $n = \frac{0.837 \ g}{171 \ g/mol} = 0.00489 \ mol$. $3$. Calculate the molarity of $Ba(OH)_2$: $M = \frac{0.00489 \ mol}{0.1 \ L} = 0.0489 \ M$. $4$. Since $Ba(OH)_2$ is a strong base,$[OH^-] = 2 	imes [Ba(OH)_2] = 2 	imes 0.0489 = 0.0978 \ M$. $5$. Calculate $pOH$: $pOH = -\log(0.0978) \approx 1.01$. $6$. Calculate $pH$: $pH = 14 - pOH = 14 - 1.01 = 12.99 \approx 12.91$ (considering rounding differences in atomic mass precision).

$0.837 \ g$ of $Ba(OH)_2$ is dissolved in $100 \ mL$ of solution. Calculate the $pH$ of the solution. (Atomic masses: $Ba = 137, O = 16, H = 1$)

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