The pH of the solution obtained by mixing 100 | vedclass

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

Question

(B) For solution $I, pH=3$ $\Rightarrow [H^{+}] = 10^{-3} \ M$ $V_{1} = 100 \ mL$ For solution $II, pH = 4$ $\Rightarrow [H^{+}] = 10^{-4} \ M$ $V_{2}

Vedclass · Accepted Answer

$4-\log 2.8$

Vedclass · Answer

(B) For solution $I, pH=3$ $\Rightarrow [H^{+}] = 10^{-3} \ M$ $V_{1} = 100 \ mL$ For solution $II, pH = 4$ $\Rightarrow [H^{+}] = 10^{-4} \ M$ $V_{2} = 400 \ mL$ Concentration of resulting solution $M = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}$ $M = \frac{10^{-3} 	imes 100 + 10^{-4} 	imes 400}{100 + 400}$ $M = \frac{0.1 + 0.04}{500} = \frac{0.14}{500} = 2.8 	imes 10^{-4} \ M$ $[H^{+}] = 2.8 	imes 10^{-4} \ M$ $pH = -\log [H^{+}] = -\log (2.8 	imes 10^{-4})$ $pH = 4 - \log 2.8$

The $pH$ of the solution obtained by mixing $100 \ mL$ of a solution of $pH=3$ with $400 \ mL$ of a solution of $pH=4$ is

Similar Questions

If the $pH$ of a solution is $2$,its normality will be

Which of the following has a $pH$ of $12$?

Which of the following statements is not correct?

The $pH$ of a $0.001\,M\,NaOH$ solution will be:

When an aqueous solution of a strong acid with $pH = 5$ is mixed with an equal volume of an aqueous solution of a strong acid with $pH = 3$,what will be the $pH$ of the resulting solution?

Vedclass Test Series

Exam Paper Generator

Online Exam Module