Chemistry

Solutions and pH Chemistry

400 Questions

Solutions and pH chemistry focus on the acidity or alkalinity of various substances, from human urine to acidic soils. Questions involve titration calculations, molarity determination, and understanding mole fractions. This topic is a staple in the chemistry sections of state and national level competitive exams.

Calculating solution molarityMole fraction calculationpH value interpretationAcidic soils and pHTitration and neutralization

Solutions and pH Chemistry Questions

Multiple choice

What is the normality of 96% solution of H2SO4 of specific gravity 1.84.

  1. 30.05 N

  2. 5.05 N

  3. 40.9 N

  4. 36.05 N

  5. 55.0 N
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Mass of 1 litre of H2SO4 solution = volume * density = 1000 *1.84 g = 1840 g = mass of H2SO4 present in 1 litre 96% H2SO4 solution = (96 / 100) *1840 g = 1766.4 g Or, strength of H2SO4 solution = 1766.4 g / litre Or Normality (N) = (Strength (gL-1) / equivalent Mass)   = (1766.4 / 49) = 36.05 (N); which match with the given option, hence this is a correct option.

Multiple choice

10 ml of (N/2) HCl, 30 ml of (N/10) HNO3 and 75 ml of (N/5) HNO3 are mixed, the normality of H+ in the resulting solution is

  1. 0.2

  2. 0.4

  3. 0.6

  4. 0.8

  5. Insufficient data
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equiv. of H+ in 30 ml of (N / 2) HCl = (10 / 2) 10^–3 The equiv. of H+ in 30 ml of (N / 10) HNO3 = (30 / 10) * 10^–3 The equiv. of H+ in 75 ml of (N / 10) HNO3 = (75 / 5) * 10^–3. Hence, total equiv. of H+ (5 + 3 + 15) * 10^–3 = 23 10^–3 Total volume of solution = 115 ml Hence, normality of H+ in the resulting mixture = [(23 * 10^–3 * 10^3) / 115) (N) = (N / 5) = 0.2 (N). Hence, this is a correct option.

Multiple choice

A 100 L sample of hard water, having only temporary hardness is dried to obtain 20 mg of CaCO3 as residue. Which of the following may be true for the sample of water? (Density = 1 g/ml)

  1. [ca+2] = 122 ppm

  2. [ca+2] = 80 ppm

  3. [ca+2] = 244 ppm

  4. [ca+2] = 180 ppm

  5. [ca+2] = 8 ppm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Mol. mass of CaCO3 = 100 Mol. mass of Ca = 40 20 mg of CaCO3 has ((40*20) / 100) = 8 mg of Ca+2 ions Since, 8 mg of Ca+2 in 100 L sample of hard water. 10*1000 ml sample of hard water has 8 mg of Ca+2 therefore, 10^6 ml sample of hard water has 80 mg of Ca+2,i.e., 80 ppm, which is the required solution of the given problem. Hence, this is a correct option.

Multiple choice

What is the mass of sodium bromate and molarity of the solution necessary to prepare 85.4 ml of 0.672 N solution when the half reaction is, BrO3– + 6H+ + 6e → Br– + 3H2O.

  1. 1.672 g, 0.112 M

  2. 2.0 g, 0.112 M

  3. 3.234 g, 0.112 M

  4. 1.92 g, 0.122 M

  5. 1.446 g, 0.122 M
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Molecular mass of NaBrO3 = 151 Each bromate ion takes up 6 electrons; therefore Eq. mass of NaBrO3 = Mol. mass / 6 = 151 / 6 Amount of NaBrO3 in 85.5 ml 0.672 N solution= (0.672 / 1000) * (151 / 6) * (85.5) g = 1.446 g. Molarity = (Normality / nf) = 0.672 / 6 = 0.112 M. Hence, this is a correct option.

Multiple choice

214.2 g of sugar syrup contains 34.2 g of sugar. Calculate the molality of the solution?

  1. 0.655 m

  2. 0.755 m

  3. 0.555 m

  4. 0.455 m

  5. 0.355 m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Mass of sugar = 34.2 g No. of moles of sugar = (34.2 / Mol. mass) = 34.2 / 342 =0.1 Mass of water = (214.2 – 43.2) = 180 g = 0.18 kg Molality (m) = (No. of moles of sugar / Mass of H2O in kg) = 0.1 / 0.18 = 0.555 m, which is the required solution of the given problem. Hence, this is a correct option.

Multiple choice

Calculate the normality of a solution obtained by mixing 100 ml (N/10) H2SO4, 50 ml (N/5) HNO3 and 25 ml (N/5) HCl solution.

  1. 0.3234

  2. 0.4453

  3. 0.545

  4. 0.2286

  5. 0.1123
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the final normality be N. Total volume = (100 + 50 + 25) ml =175 ml So, 175 * N = N1V1 + N2V2 + N3V3------(1) Where, N1, N2, N3 are the normalities of H2SO4, HNO3 and HCl respectively. And V1, V2, V3 are the volumes of H2SO4, HNO3 and HCl respectively. Now, we have, 175 * N = ((100)(1 / 10)) + ((50)(1 / 2)) + ((25)* (1 / 5) = 40 therefore, N = 40 / 175  [where, N = normality of the mixture solution] = 0.2286, which is the required solution of the given problem, hence this is a correct option.

Multiple choice

Isoelectric pH of leucine is

  1. 5

  2. 5.2

  3. 5.8

  4. 6.0
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The isoelectric pH (pI) of leucine is approximately 6.0. This is calculated as the average of its carboxyl group pKa (~2.4) and amino group pKa (~9.6), giving pI = (2.4 + 9.6)/2 = 6.0. At this pH, leucine exists as a zwitterion with no net charge.

Multiple choice

The normal Ph of the human blood is _____.

  1. 7.45

  2. 7.34

  3. 7.23

    1. 12
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The normal pH of human arterial blood is 7.35-7.45, with 7.40 being ideal. A pH of 7.45 is within normal range, though at the upper end. pH below 7.35 indicates acidosis, above 7.45 indicates alkalosis. The body tightly regulates blood pH through respiratory and renal mechanisms.