Equal volumes of three acid solutions of pH 3 | vedclass

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

Question

(D) The $pH$ is defined as $pH = -\log[H^{+}]$,which implies $[H^{+}] = 10^{-pH}$.For the three solutions:$[H^{+}]_1 = 10^{-3} \ M$$[H^{+}]_2 = 10^{-4

Vedclass · Accepted Answer

$3.7$

Vedclass · Answer

(D) The $pH$ is defined as $pH = -\log[H^{+}]$,which implies $[H^{+}] = 10^{-pH}$.For the three solutions:$[H^{+}]_1 = 10^{-3} \ M$$[H^{+}]_2 = 10^{-4} \ M$$[H^{+}]_3 = 10^{-5} \ M$Assuming equal volumes $V$ for each solution,the total volume of the mixture is $3V$.The total moles of $H^{+}$ ions are $n_{total} = (10^{-3} 	imes V) + (10^{-4} 	imes V) + (10^{-5} 	imes V) = V(10^{-3} + 0.1 	imes 10^{-3} + 0.01 	imes 10^{-3}) = V(1.11 	imes 10^{-3})$.The final concentration is $[H^{+}]_{mix} = \frac{n_{total}}{3V} = \frac{1.11 	imes 10^{-3}}{3} = 0.37 	imes 10^{-3} \ M$.Converting this to the form $	imes 10^{-4} \ M$,we get $3.7 	imes 10^{-4} \ M$.

Equal volumes of three acid solutions of $pH$ $3$,$4$ and $5$ are mixed in a vessel. What will be the $H^{+}$ ion concentration in the mixture in terms of $\times 10^{-4} \ M$?

Similar Questions

$40 \, mg$ of pure sodium hydroxide is dissolved in $10 \, L$ of distilled water. The $pH$ of the solution is

The hydrogen ion concentration of $0.001 \ M$ $NaOH$ solution is

If $pH$ of $A$,$B$,$C$ and $D$ are $9.5$,$2.5$,$3.5$ and $5.5$ respectively,then the strongest acid is

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?

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module