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AP EAMCET · Chemistry · Ionic Equilibrium

The solubility of barium phosphate of molar mass ' M ' g \(\mathrm{mol}^{-1}\) in water is \(x \mathrm{~g}\) per 100 mL at 298 K . Its solubility product is \(1.08 \times\left(\frac{x}{M}\right)^a \times(10)^{\mathrm{b}}\). The values of \(a\) and \(b\) respectively are

  1. A 7,5
  2. B 5,7
  3. C 5,5
  4. D 7,7
Verified Solution

Answer & Solution

Correct Answer

(B) 5,7

Step-by-step Solution

Detailed explanation

\(\mathrm{Ba}_3\left(\mathrm{PO}_4\right)_2 \longrightarrow 3 \mathrm{Ba}^{2+}+2 \mathrm{PO}_4{ }^{3-}\)…