back to directory
NEET CHEMISTRYEquilibriumEasy

Question

4 gm of NaOH is dissolved in 1000 ml of water. The H+\text{H}^+ ion concentration will be:

A

10−1 M10^{-1} \text{ M}

B

10−13 M10^{-13} \text{ M}

C

10−4 M10^{-4} \text{ M}

D

10−10 M10^{-10} \text{ M}

Step-by-Step Solution

First, we calculate the molarity of the NaOH\text{NaOH} solution. Molar mass of NaOH=23+16+1=40 g/mol\text{NaOH} = 23 + 16 + 1 = 40 \text{ g/mol}. Number of moles of NaOH=Given massMolar mass=4 g40 g/mol=0.1 mol\text{NaOH} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{4 \text{ g}}{40 \text{ g/mol}} = 0.1 \text{ mol}. Volume of solution =1000 ml=1 L= 1000 \text{ ml} = 1 \text{ L}. Molarity (MM) =Moles of soluteVolume of solution in L=0.11=0.1 M= \frac{\text{Moles of solute}}{\text{Volume of solution in L}} = \frac{0.1}{1} = 0.1 \text{ M}. Since NaOH\text{NaOH} is a strong base, it completely dissociates into Na+\text{Na}^+ and OH−\text{OH}^- ions in aqueous solution . Therefore, the concentration of hydroxyl ions, [OH−]=0.1 M=10−1 M[\text{OH}^-] = 0.1 \text{ M} = 10^{-1} \text{ M}. We know the ionic product of water, Kw=[H+][OH−]=10−14K_w = [\text{H}^+][\text{OH}^-] = 10^{-14} at 298 K298 \text{ K} . Thus, [H+]=Kw[OH−]=10−1410−1=10−13 M[\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{10^{-14}}{10^{-1}} = 10^{-13} \text{ M}. Hence, the H+\text{H}^+ ion concentration is 10−13 M10^{-13} \text{ M}.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Equilibrium. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYEquilibriumdissolvedconcentration

More Equilibrium Questions

View all

Consider the following reaction: $\text{A}_2(g) + \text{B}_2(g) \rightleftharpoons 2\text{AB}(g)$. At equilibrium, the concentrations of $[\text{A}_2] = 3.0 \times 10^{–3} \text{ M}$; $[\text{B}_2] = 4.2 \times 10^{–3} \text{ M}$ and $[\text{AB}] = 2.8 \times 10^{–3} \text{ M}$. The value of $K_c$ for the above-given reaction in a sealed container at $527^\circ\text{C}$ is:

A.3.9
B.0.6
C.4.5
D.2
MediumSolve

Amongst the given options, which of the following molecules/ions acts as a Lewis acid?

A.$\text{OH}^-$
B.$\text{NH}_3$
C.$\text{H}_2\text{O}$
D.$\text{BF}_3$
EasySolve

Boric acid is an acid because its molecule

A.contains replaceable H⁺ ion
B.gives up a proton
C.accepts OH⁻ from water releasing proton
D.combines with proton from water molecule
MediumSolve

The following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations. pH of which one of them will be equal to 1?

A.$60 \text{ mL } \frac{M}{10} \text{ HCl } + 40 \text{ mL } \frac{M}{10} \text{ NaOH}$
B.$55 \text{ mL } \frac{M}{10} \text{ HCl } + 45 \text{ mL } \frac{M}{10} \text{ NaOH}$
C.$75 \text{ mL } \frac{M}{5} \text{ HCl } + 25 \text{ mL } \frac{M}{5} \text{ NaOH}$
D.$100 \text{ mL } \frac{M}{10} \text{ HCl } + 100 \text{ mL } \frac{M}{10} \text{ NaOH}$
MediumSolve

The tendency of $BF_3$, $BCl_3$ and $BBr_3$ to behave as Lewis acid decreases in the sequence:

A.$BCl_3 > BF_3 > BBr_3$
B.$BBr_3 > BCl_3 > BF_3$
C.$BBr_3 > BF_3 > BCl_3$
D.$BF_3 > BCl_3 > BBr_3$
MediumSolve

A compound $\text{BA}_2$ has $K_{sp} = 4 \times 10^{-12}$. Solubility of this compound will be:

A.$10^{-3} \text{ mol/L}$
B.$10^{-4} \text{ mol/L}$
C.$10^{-5} \text{ mol/L}$
D.$10^{-6} \text{ mol/L}$
EasySolve

What is the molarity of the saturated solution if the solubility product for a salt of type AB is $4 \times 10^{-8}$?

A.$2 \times 10^{-4} \text{ mol/L}$
B.$16 \times 10^{-16} \text{ mol/L}$
C.$2 \times 10^{-16} \text{ mol/L}$
D.$4 \times 10^{-4} \text{ mol/L}$
EasySolve

In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains $\text{Ag}^+$ and $\text{Pb}^{2+}$ at a concentration of $0.10 \text{ M}$. Aqueous $\text{HCl}$ is added to this solution until the $\text{Cl}^-$ concentration is $0.10 \text{ M}$. What will the concentration of $\text{Ag}^+$ and $\text{Pb}^{2+}$ at equilibrium? ($K_{sp}$ for $\text{AgCl} = 1.8 \times 10^{-10}$, $K_{sp}$ for $\text{PbCl}_2 = 1.7 \times 10^{-5}$)

A.$[\text{Ag}^+] = 1.8 \times 10^{-11} \text{ M}; [\text{Pb}^{2+}] = 1.7 \times 10^{-4} \text{ M}$
B.$[\text{Ag}^+] = 1.8 \times 10^{-7} \text{ M}; [\text{Pb}^{2+}] = 1.7 \times 10^{-6} \text{ M}$
C.$[\text{Ag}^+] = 1.8 \times 10^{-11} \text{ M}; [\text{Pb}^{2+}] = 8.5 \times 10^{-5} \text{ M}$
D.$[\text{Ag}^+] = 1.8 \times 10^{-9} \text{ M}; [\text{Pb}^{2+}] = 1.7 \times 10^{-3} \text{ M}$
MediumSolve

Why 32,000+ students choose TopperSquare

Everything you need to crack NEET

15,000+ Chapter MCQs

Physics, Chemistry, Biology — chapter-wise, topic-wise practice

AI Performance Analytics

Know your weak topics. Get a personalised study plan.

Full NEET Mock Tests

180 questions, timed, with detailed score analysis

PYQ Sets 2010–2024

15 years of past papers with solved explanations

Start Practising Free

No credit card · 30-second signup · Free forever plan included

This neet chemistry practice question is part of the TopperSquare free question bank. TopperSquare offers 15,000+ chapter-wise NEET MCQs across Physics, Chemistry, and Biology with detailed step-by-step explanations, full mock tests, NEET PYQs (2010–2024), and an AI-powered performance analytics dashboard. browse all neet practice questions → · practice chemistry sets →

Explore more subjects:Physics MCQs•Chemistry MCQs•Biology MCQs•Botany MCQs•Zoology MCQs•NEET Mock Tests•NEET PYQs