Calculate the volume of 0.1 M NaOH required | vedclass

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(A) $1$. Given $pH = 2.25$ for $HCl$. The concentration of $H^+$ ions is $[H^+] = 10^{-pH} = 10^{-2.25}$.$2$. $[H^+] = 10^{0.75} 	imes 10^{-3} \appro

Vedclass · Accepted Answer

$16.9$

Vedclass · Answer

(A) $1$. Given $pH = 2.25$ for $HCl$. The concentration of $H^+$ ions is $[H^+] = 10^{-pH} = 10^{-2.25}$.$2$. $[H^+] = 10^{0.75} 	imes 10^{-3} \approx 5.623 	imes 10^{-3} \ M$.$3$. Since $HCl$ is a strong acid,$[HCl] = [H^+] = 5.623 	imes 10^{-3} \ M$.$4$. The neutralization reaction is $HCl + NaOH ightarrow NaCl + H_2O$. The stoichiometry is $1:1$,so $n_{HCl} = n_{NaOH}$.$5$. $n_{HCl} = M_{HCl} 	imes V_{HCl} = (5.623 	imes 10^{-3} \ M) 	imes (0.300 \ L) = 1.687 	imes 10^{-3} \ mol$.$6$. $V_{NaOH} = \frac{n_{NaOH}}{M_{NaOH}} = \frac{1.687 	imes 10^{-3} \ mol}{0.1 \ M} = 0.01687 \ L = 16.87 \ mL \approx 16.9 \ mL$.

Calculate the volume of $0.1 \ M \ NaOH$ required to complete the neutralization of $300 \ mL$ of $HCl$ solution having $pH = 2.25$. (in $mL$)

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