The pH of a solution obtained by mixing 50 , | vedclass

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(B) The reaction is: $HCl_{(aq)} + NaOH_{(aq)} ightarrow NaCl_{(aq)} + H_2O_{(\ell)}$Initial moles of $HCl = 50 \, mL 	imes 1 \, M = 50 \, mmol$.In

Vedclass · Accepted Answer

$6021$

Vedclass · Answer

(B) The reaction is: $HCl_{(aq)} + NaOH_{(aq)} ightarrow NaCl_{(aq)} + H_2O_{(\ell)}$Initial moles of $HCl = 50 \, mL 	imes 1 \, M = 50 \, mmol$.Initial moles of $NaOH = 30 \, mL 	imes 1 \, M = 30 \, mmol$.After reaction,$HCl$ remaining $= 50 - 30 = 20 \, mmol$.Total volume of the solution $= 50 \, mL + 30 \, mL = 80 \, mL$.Concentration of $[H^+] = [HCl] = \frac{20 \, mmol}{80 \, mL} = 0.25 \, M = 2.5 	imes 10^{-1} \, M$.$pH = -\log[H^+] = -\log(2.5 	imes 10^{-1}) = -(\log 2.5 - 1) = 1 - 0.3979 = 0.6021$.Given $pH = x 	imes 10^{-4}$,so $0.6021 = x 	imes 10^{-4}$.$x = 0.6021 	imes 10^4 = 6021$.

The $pH$ of a solution obtained by mixing $50 \, mL$ of $1 \, M \, HCl$ and $30 \, mL$ of $1 \, M \, NaOH$ is $x \times 10^{-4}$. The value of $x$ is ...... . (Nearest integer) $[\log 2.5 = 0.3979]$

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