0.1 mL of 0.001 M HCl solution is diluted w | vedclass

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(A) Step $1$: Calculate the initial moles of $H^+$. $n = M 	imes V = 0.001 \ mol/L 	imes 0.0001 \ L = 10^{-7} \ mol$.Step $2$: Calculate the final c

Vedclass · Accepted Answer

$6.96$

Vedclass · Answer

(A) Step $1$: Calculate the initial moles of $H^+$. $n = M 	imes V = 0.001 \ mol/L 	imes 0.0001 \ L = 10^{-7} \ mol$.Step $2$: Calculate the final concentration of $H^+$ in $10 \ L$ solution. $[H^+] = \frac{10^{-7} \ mol}{10 \ L} = 10^{-8} \ M$.Step $3$: Since the concentration of $H^+$ from $HCl$ is very low $(10^{-8} \ M)$,we must consider the contribution of $H^+$ from water $(10^{-7} \ M)$.Total $[H^+] = 10^{-8} + 10^{-7} = 1.1 	imes 10^{-7} \ M$.Step $4$: Calculate $pH = -\log[H^+] = -\log(1.1 	imes 10^{-7}) = 7 - \log(1.1) \approx 7 - 0.04 = 6.96$.

$0.1 \ mL$ of $0.001 \ M$ $HCl$ solution is diluted with water to form $10 \ L$. Calculate the $pH$ of the dilute solution.

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