Chemistry

Electrochemistry and Solutions

206 Questions

Work through advanced chemistry questions focusing on electrochemistry and properties of solutions. Key themes include electrolytic conductivity, colloidal solutions, and solubility products. These concepts are frequently tested in JEE advanced, NEET, and university level chemistry competitive examinations.

Colloidal solutionsMolar conductivitySolubility productElectrolytic conductivityHydrolysis of saltsEquivalent conductivity

Electrochemistry and Solutions Questions

Multiple choice

The figure illustrates the conductmetric titration of hydrochloric acid against sodium hydroxide. The conductance at B is due to

  1. Sodium and chloride ions

  2. Sodium and hydroxide ions

  3. Hydrogen and chloride ions

  4. H2O molecules
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In conductometric titration of HCl against NaOH, at the equivalence point (B) and beyond, the solution contains Na+ and Cl- ions (from the neutralization reaction) and excess OH- ions. The conductance at B is primarily due to Na+ and Cl- ions present as products of the neutralization. Initially, H+ and Cl- contribute; after neutralization, Na+ and Cl- remain.

Multiple choice

Molecular formula of Glauber’s salt is _________.

  1. Na2SO4.10.H2O

  2. Na(NH4)HPO4

  3. Na2CO3.10H2O

  4. Na2SO4

  5. Na2SO4.7H2O
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Molecular formula of Glauber’s salt is Na2SO4.10.H2O. Dissolving salt cake (Na2SO4) in water and crystallisation at 32 oC yields Glauber’s salt.

Multiple choice

The solubility of Na2SO4 ______________ with increase in temperature upto 305K and thereafter _____________.

  1. increases, decreases

  2. decreases, increases

  3. increases, remains constant

  4. remains constant, decreases

  5. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The initial solubility is due to the hydrated salt. Below 305K, Na2SO4.10H2O crystalises out. Above 305K the anhydrous salt separates out due to its negative temperature co-efficient of solubility.

Multiple choice

Consider a solution saturated with Ag2SO4(s) at equilibrium Ag2SO4(s) ⇌ 2Ag+(aq) + SO42-(aq) On addition of NaCl(s), the amount of Ag2SO4 in equilibrium will

  1. increase

  2. decrease

  3. first decrease then increase

  4. remains constant.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This option is correct and can be explained as follows: When NaCl(s) is added to the system it dissociates as NaCl(s)  ⇌  Na+(aq)  +  Cl-(aq)The Cl- ions obtained from the dissociation of solid NaCl will combine with Ag+ ions to precipitate AgCl. Hence the equilibrium will be shifted in forward direction in accordance with Le-Chatelier's principle and more Ag+ and SO42- ions will be formed resulting in consumption of Ag2SO4 and hence its concentration will decrease.

Multiple choice

When an equimolar mixture of Cu2S and CuS is titrated with Ba(MnO4) in acidic medium, the final products contain Cu2+, SO2 and Mn2+. If the mol. wt. of Cu2S, CuS and Ba(MnO4)2 are M1, M2, M3 respectively, then,

  1. eq. wt. of Cu2S is M1/8

  2. eq. wt. of CuS is M2/4

  3. eq. wt. of Ba(MnO4)2 is M3/5

  4. Cu2S and CuS, both have same equivalent in mixture

  5. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The required reaction is, MnO4– +  8H+ +5e  →  Mn2+ + 4H2O Cu+ – e → Cu2+ S2– – 6e → S4+ Therefore, eq. wt. of Cu2S = (M1 / ((1*2) + 6)) = M1 / 8 Hence, this is a correct option.  

Multiple choice

The amount of 0.02 M AgNO3 volume required for complete precipitation of chloride ions present in 30 mL of 0.02 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is close to:

  1. 4 mL

  2. 5 mL

  3. 6 mL

  4. 7 mL

  5. 8 mL
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

[Cr(H2O)5Cl]Cl2 + 2AgNO3 → 2AgCl + [Cr(H2O)5Cl](NO32
Number of ionisable chloride ions in complex [Cr(H2O)5Cl]Cl2 = 2 Millimoles = Molarity x Volume (mL) x 2 = 0.02 x 30 x 2 = 1.2 Therefore, required Ag+ ions = 1.2 millimoles According to Millimoles = Molarity x V (mL) 1.2 = 0.2 x VV = 6 mL

Multiple choice

According to conductance measurement, how many ions are present in a dilute solution of CoCl3.4NH3?

  1. 2

  2. 3

  3. 5

  4. 6
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

CoCl3.4NH3 dissociates in water to give [Co(NH3)4Cl2]+ and Cl- ions, totaling 2 ions per formula unit. Conductance measurements directly count the number of ions produced, not the coordination number. The compound exists as [Co(NH3)4Cl2]Cl, where the complex cation and one chloride ion separate in solution.

Multiple choice

The volume (in mL) of 0.2 M AgNO3 required for complete precipitation of chloride ions present in 50 mL of 0.02 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is __________.

  1. 1 mL

  2. 2 mL

  3. 5 mL

  4. 10 mL

  5. 20 mL
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

[Cr(H2O)5Cl]Cl2 + 2AgNO3 → 2AgCl + [Cr(H2O)5Cl](NO32 Number of ionisable chloride ions in complex [Cr(H2O)5Cl]Cl2 = 2. Millimoles = Molarity x Volume (mL) x 2= 0.02 x 50 x 2 = 2.0 Therefore, required Ag + ions = 2.0 millimoles Millimoles = Molarity x V (mL) 2.0 = 0.2 x V V = 10 mL

Multiple choice

Melting point of NaCl is ______.

  1. 800°C

  2. 825°C

  3. 850°C

  4. None of these

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Sodium chloride (NaCl) is a common ionic compound with a high melting point, specifically around 801 degrees Celsius.

Multiple choice

Chlorine gas (8 mg/L as Cl2) was added to a drinking water sample. If the free chlorine residual and pH was measured to be 2 mg/L (as Cl2) and 7.5 respectively, what is the concentration of residual OCF ions in the water? Assume that the chlorine gas added to the water is completely converted to HOCI and OCI–. Atomic Weight of Cl is 35.5. Given: OCl– + H+$\underline{\underrightarrow{K}}$HOCl, K = 107.5

  1. 1.408x10–5 moles/L

  2. 2.817x10–5 moles/L

  3. 5.634x10–5 moles/L

  4. 1.127x10–4 moles/L
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation