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(B) Number of milliequivalents $(meq)$ of $Ca(OH)_2 = 	ext{Volume} 	imes 	ext{Molarity} 	imes n	ext{-factor} = 50 \ mL 	imes 0.01 \ M 	imes 2 =

Vedclass · Accepted Answer

$-14.275 \ J$

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(B) Number of milliequivalents $(meq)$ of $Ca(OH)_2 = 	ext{Volume} 	imes 	ext{Molarity} 	imes n	ext{-factor} = 50 \ mL 	imes 0.01 \ M 	imes 2 = 1.00 \ meq$. Number of $meq$ of $HCl = 25 \ mL 	imes 0.01 \ M 	imes 1 = 0.25 \ meq$. Since $HCl$ is the limiting reagent,$0.25 \ meq$ of $HCl$ will be neutralized by $0.25 \ meq$ of $Ca(OH)_2$. Enthalpy of neutralization for $1 \ eq$ of $H^+$ and $OH^-$ is $-57.1 \ kJ$. For $0.25 \ meq$ $(0.25 	imes 10^{-3} \ eq)$,the enthalpy change is: $\Delta H = 0.25 	imes 10^{-3} \ eq 	imes (-57.1 \ kJ \ eq^{-1}) = -14.275 	imes 10^{-3} \ kJ = -14.275 \ J$.

Calculate the enthalpy change when $50 \ mL$ of $0.01 \ M$ $Ca(OH)_2$ reacts with $25 \ mL$ of $0.01 \ M$ $HCl$. Given that $\Delta H^o_{\text{neutralisation}}$ of strong acid and strong base is $-57.1 \ kJ \ mol^{-1}$. [Assuming that $Ca(OH)_2$ is a strong base]

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