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(B) The balanced chemical equation for the titration is:$Ba(OH)_{2(aq)} + 2HCl_{(aq)} \rightarrow BaCl_{2(aq)} + 2H_2O_{(\ell)}$Calculate the millimol

Vedclass · Accepted Answer

$1825$

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(B) The balanced chemical equation for the titration is:$Ba(OH)_{2(aq)} + 2HCl_{(aq)} ightarrow BaCl_{2(aq)} + 2H_2O_{(\ell)}$Calculate the millimoles of $Ba(OH)_2$ used:$n_{Ba(OH)_2} = M 	imes V_{(mL)} = 0.1 \ M 	imes 25.0 \ mL = 2.5 \ mmol$From the stoichiometry, $1 \ mol$ of $Ba(OH)_2$ reacts with $2 \ mol$ of $HCl$:$n_{HCl} = 2 	imes n_{Ba(OH)_2} = 2 	imes 2.5 \ mmol = 5.0 \ mmol$Calculate the mass of $HCl$ in $mg$:$Mass = n 	imes Molar \ mass = 5.0 \ mmol 	imes 36.5 \ mg/mmol = 182.5 \ mg$Given $x \ mg = 182.5 \ mg$, we need to express this as $x 	imes 10^{-1}$:$182.5 = 1825 	imes 10^{-1}$Therefore, $x = 1825$.

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