200 mL of an aqueous solution of HCl (pH=2) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For $HCl$ solution: $pH = 2$,so $[H^+] = 10^{-2} \ M$. Moles of $H^+ = 10^{-2} \ mol/L 	imes 0.2 \ L = 2 	imes 10^{-3} \ mol$. For $NaOH$ soluti

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(B) For $HCl$ solution: $pH = 2$,so $[H^+] = 10^{-2} \ M$. Moles of $H^+ = 10^{-2} \ mol/L 	imes 0.2 \ L = 2 	imes 10^{-3} \ mol$. For $NaOH$ solution: $pH = 12$,so $pOH = 14 - 12 = 2$. $[OH^-] = 10^{-2} \ M$. Moles of $OH^- = 10^{-2} \ mol/L 	imes 0.3 \ L = 3 	imes 10^{-3} \ mol$. Since $n(OH^-) > n(H^+)$,the reaction $H^+ + OH^- ightarrow H_2O$ occurs. Remaining moles of $OH^- = 3 	imes 10^{-3} - 2 	imes 10^{-3} = 1 	imes 10^{-3} \ mol$. The total volume of the resulting solution is $1.0 \ L$. Concentration of $OH^- = \frac{1 	imes 10^{-3} \ mol}{1.0 \ L} = 10^{-3} \ M$. $pOH = -\log(10^{-3}) = 3$. $pH = 14 - pOH = 14 - 3 = 11$.

$200 \ mL$ of an aqueous solution of $HCl$ $(pH=2)$ is mixed with $300 \ mL$ of aqueous solution of $NaOH$ $(pH=12)$ and is diluted to $1.0 \ L$. The $pH$ of the resulting solution is

Similar Questions

Which of the following will have the highest $pH$?

What is the $pH$ of the $NaOH$ solution when $0.04 \ g$ of it is dissolved in water to make a $100 \ mL$ solution?

$pH$ of $10^{-8} \ M$ $Ba(OH)_2$ solution will be

If $10^{-6} \ M \ HCl$ is diluted $100$ times,its $pH$ will be:

An aqueous solution of $HCl$ with $pH \ 1.0$ is diluted by adding an equal volume of water (ignoring the dissociation of water). The $pH$ of the $HCl$ solution would be (Given $\log 2 = 0.30$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module