Chemistry Class 11 NCERT Solutions: Chapter 7 Equilibrium Part 27

Chemistry Class 11 NCERT Solutions Chapter 7 Equilibrium Part 27

Q: 68. The solubility product constant of Equation and Equation are Equation respectively. Calculate the ratio of the molarities of their saturated solutions.

Answer:

Let s be the solubility of Equation

Then, Equation

Equation

Equation

Equation

Let Equation be the solubility of Equation

Equation

Equation

Equation

Therefore, the ratio of the molarities of their saturated solution is

Equation

Q: 69. Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate Equation

Answer:

When equal volumes of sodium iodate and cupric chlorate solutions are mixed together, then the molar concentrations of both solutions are reduced to half i.e., 0.001 M. Then,

Equation

Now, the solubility equilibrium for copper iodate can be written as:

Equation

Ionic product of copper iodate:

Equation

Equation

Equation

Since the ionic product Equation is less than Equation ), precipitation will not occur.

Q: 70. The ionization constant of benzoic acid is Equation and Equation for silver benzoate is Equation How many times is silver benzoate more soluble in a buffer of Equation compared to its solubility in pure water?

Answer:

Since Equation

Equation

Equation

Equation

Equation

Let the solubility of Equation be x mol/L.

Then,

Equation

Equation

Equation

Equation

Equation

Equation

Equation

Thus, the solubility of silver benzoate in a pH 3.19 solution is 1.66 × 10–6 mol/L. Now, let the solubility of C6H5COOAg be x’mol/L.

Then, Equation

Equation

Equation

Equation

Equation

Hence, Equation is approximately 3.317 times more soluble in a low pH solution.

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