If 1 mL of 1 M HCl solution is added to 99 | vedclass

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(B) Initially,the $NaCl$ solution is neutral,so its $pH = 7$. After adding $1 \ mL$ of $1 \ M \ HCl$ to $99 \ mL$ of $NaCl$ solution,the total volume

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) Initially,the $NaCl$ solution is neutral,so its $pH = 7$. After adding $1 \ mL$ of $1 \ M \ HCl$ to $99 \ mL$ of $NaCl$ solution,the total volume becomes $100 \ mL$. The concentration of $H^+$ ions in the new solution is $[H^+] = \frac{n}{V} = \frac{1 \ mL 	imes 1 \ M}{100 \ mL} = 0.01 \ M = 10^{-2} \ M$. The new $pH$ of the solution is $pH = -\log[H^+] = -\log(10^{-2}) = 2$. The change in $pH$ is $X = |pH_{initial} - pH_{final}| = |7 - 2| = 5$ units.

If $1 \ mL$ of $1 \ M \ HCl$ solution is added to $99 \ mL$ of aqueous solution of $NaCl$,the $pH$ of the solution changes by $X$ units. The value of $X$ is:

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