(C) Initial concentration of $[H^{+}] = 1 \times 10^{-4} \ M$. When the solution is diluted with an equal volume of water,the total volume becomes dou

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

(C) Initial concentration of $[H^{+}] = 1 \times 10^{-4} \ M$. When the solution is diluted with an equal volume of water,the total volume becomes double,so the concentration of $[H^{+}]$ is halved: $[H^{+}]_{new} = \frac{1 \times 10^{-4}}{2} = 0.5 \times 10^{-4} \ M$. Using the ionic product of water,$K_w = [H^{+}][OH^{-}] = 1 \times 10^{-14}$. $[OH^{-}] = \frac{K_w}{[H^{+}]_{new}} = \frac{1 \times 10^{-14}}{0.5 \times 10^{-4}} = 2 \times 10^{-10} \ mol \ dm^{-3}$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Mix Examples-Ionic Equilibrium

Medium

Hydrogen ion concentration of an aqueous solution is $1 \times 10^{-4} \ M$. The solution is diluted with an equal volume of water. The hydroxyl ion concentration of the resultant solution in terms of $mol \ dm^{-3}$ is

A
$1 \times 10^{-8}$
B
$1 \times 10^{-6}$
C
$2 \times 10^{-10}$
D
$0.5 \times 10^{-10}$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Chemistry Questions

Chapter

6-2.Equilibrium-II (Ionic Equilibrium)

Subtopic

Mix Examples-Ionic Equilibrium

$A$ $20 \ mL$ sample of a $0.1 \ M$ solution of the weak monoprotic acid $HA$ is titrated with $0.25 \ M \ NaOH$. The volume of $NaOH$ solution required to reach the equivalence point is: ............. $mL$

Medium

View Solution

Assign $A, B, C, D$ from the given type of reaction.
$CH_3COOAg(s) + HNO_3(aq) \longrightarrow AgNO_3(aq) + CH_3COOH(aq)$

Medium

View Solution

Estimate the approximate $pK_a$ of $0.5 \ M$ $CH_3COOH$. Degree of dissociation (ionisation) is $0.15$ $(\log 1.32 = 0.12)$.

DifficultTS EAMCET 2018

View Solution

$50 \ mL$ of $10^{-3} \ M \ HCl$ is mixed with $50 \ mL$ of $0.05 \ M \ CH_3COOH$ $(pK_a = 4.7)$. The degree of dissociation of water in the resulting solution is

Medium

View Solution

When $10 \, mL$ of $0.1 \, M$ acetic acid $(pK_a = 5.0)$ is titrated against $10 \, mL$ of $0.1 \, M$ ammonia solution $(pK_b = 5.0)$,the equivalence point occurs at $pH$:

MediumAIIMS 2005

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Hydrogen ion concentration of an aqueous solution is $1 \times 10^{-4} \ M$. The solution is diluted with an equal volume of water. The hydroxyl ion concentration of the resultant solution in terms of $mol \ dm^{-3}$ is

Similar Questions

$A$ $20 \ mL$ sample of a $0.1 \ M$ solution of the weak monoprotic acid $HA$ is titrated with $0.25 \ M \ NaOH$. The volume of $NaOH$ solution required to reach the equivalence point is: ............. $mL$

Assign $A, B, C, D$ from the given type of reaction.$CH_3COOAg(s) + HNO_3(aq) \longrightarrow AgNO_3(aq) + CH_3COOH(aq)$

Estimate the approximate $pK_a$ of $0.5 \ M$ $CH_3COOH$. Degree of dissociation (ionisation) is $0.15$ $(\log 1.32 = 0.12)$.

$50 \ mL$ of $10^{-3} \ M \ HCl$ is mixed with $50 \ mL$ of $0.05 \ M \ CH_3COOH$ $(pK_a = 4.7)$. The degree of dissociation of water in the resulting solution is

When $10 \, mL$ of $0.1 \, M$ acetic acid $(pK_a = 5.0)$ is titrated against $10 \, mL$ of $0.1 \, M$ ammonia solution $(pK_b = 5.0)$,the equivalence point occurs at $pH$:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Assign $A, B, C, D$ from the given type of reaction.
$CH_3COOAg(s) + HNO_3(aq) \longrightarrow AgNO_3(aq) + CH_3COOH(aq)$