What is the pH of the resulting solution when | vedclass

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

Question

(D) Let the volume of each solution be $V \ L$. Total volume of the mixture $= 2V \ L$. Moles of $NaOH = 0.1 	imes V = 0.1V$. Moles of $HCl = 0.01

Vedclass · Accepted Answer

$12.65$

Vedclass · Answer

(D) Let the volume of each solution be $V \ L$. Total volume of the mixture $= 2V \ L$. Moles of $NaOH = 0.1 	imes V = 0.1V$. Moles of $HCl = 0.01 	imes V = 0.01V$. Since $NaOH$ is a strong base and $HCl$ is a strong acid,they react as: $NaOH + HCl ightarrow NaCl + H_2O$. Moles of $NaOH$ left unneutralized $= 0.1V - 0.01V = 0.09V$. Concentration of $[OH^-] = \frac{0.09V}{2V} = 0.045 \ M$. $pOH = -\log(0.045) = 1.3468 \approx 1.35$. $pH = 14 - pOH = 14 - 1.35 = 12.65$.

What is the $pH$ of the resulting solution when equal volumes of $0.1 \ M \ NaOH$ and $0.01 \ M \ HCl$ are mixed?

Similar Questions

The $pH$ value of a $\frac{N}{10} \ NaOH$ solution is:

$2.8 \text{ g}$ of $KOH$ is dissolved in a $500 \text{ mL}$ solution at $298 \text{ K}$. What is the $pH$ of the solution? (Molar mass of $KOH = 56 \text{ g/mol}$)

What is the $pH$ value for $\frac{N}{10} \ H_2SO_4$?

The $pH$ value of $0.1 \, M \, HCl$ is approximately $1$. What will be the approximate $pH$ value of $0.05 \, M \, H_2SO_4$?

What is the $pH$ of a $10^{-4} \ M \ OH^{-}$ solution at $330 \ K$,if $K_w$ at $330 \ K$ is $10^{-13.6}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module