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Equilibrium — Chemistry STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceChemistryEquilibrium5 Marks
Question
  1. Define solubility product. Write solubility product expression for Zr3(PO4)4.
  2. Calculate the pH of 0.01 M CH3COOH solution. [Ka(CH2COOH) = 1.74 × 10-51
  3. Explain why NaCl is precipitated when HCl(g) is passed through the saturated solution of NaCl.
✓

Answer

  1. Solubility Product: It is defined as the product of molar concentrations of the ions (formed in the saturated solution at a given temperature) raised to the power equal to the number of times each ion occurs in the equation for solubility equilibrium,

 $\text{Zr}_3(\text{PO}_4)_4\rightleftharpoons3\text{Zr}^{4+}+4\text{PO}^{3-}_4$

$\text{K}_{\text{sp}}=[\text{Zr}^{4+}][\text{PO}^{3-}_4]^4$

  1. $\text{pH}=-\log[\text{H}_3\text{O}^+]$

$=-\log\text{C}\alpha$

$=-\log\sqrt{\text{K}_\text{a}\times\text{C}}$

$=-\log\sqrt{1.74\times10^{-5}\times0.01}$

$=-\log\sqrt{1.74\times10^{-7}}$

$=-\log\sqrt{17.4\times10^{-8}}$

$=-\log4.17\times10^{-4}$

$=-\log4.17-\log10^{-4}$

$=-0.6217+4.000$

$=3.3783$

  1. It is due to common ion, CI- increase, therefore rate of backward reaction increases, solubility of NaCl decreases.

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