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Equilibrium — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryEquilibrium5 Marks
Question
  1. Define solubility product. Write solubility product expression for $Zr_3(PO_4)_4$.
  2. Calculate the pH of $0.01 M CH_3COOH$ solution. $[K_a(CH_2COOH) = 1.74 \times 10^{-51}$
  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,
  1. $=-\log\text{C}\alpha$
    $=-\log\sqrt{\text{K}_\text{a}\times\text{C}}$
    $=-\log\sqrt{1.74\times10^{-5}\times0.01}$
    $=-\log\sqrt{1.74\times10^{-7}}$
    $=-\log\sqrt{17.4\times10^{-8}}$
    $=-\log4.17\times10^{-4}$
    $=-\log4.17-\log10^{-4}$
    $=-0.6217+4.000$
    $=3.3783$
  2. It is due to common ion, $CI^-$ increase, therefore rate of backward reaction increases, solubility of NaCl decreases.
  3. $\text{Zr}_3(\text{PO}_4)_4\rightleftharpoons3\text{Zr}^{4+}+4\text{PO}^{3-}_4$
    $\text{K}_{\text{sp}}=[\text{Zr}^{4+}][\text{PO}^{3-}_4]^4$
  4. $\text{pH}=-\log[\text{H}_3\text{O}^+]$

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