The $K_{s p}$ of $A g_{2} C r O_{4}$ is $1.1 \times 10^{- 12}$ at 298K. The solubility (in mol/L) of $A g_{2} C r O_{4}$ in 0.1M $A g N O_{3}$ solution is:

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0.01~M solution of an acid HA freezes at −0.0205∘C. If Kf for water is 1.86 Kkgmol−1.The ionization constant for the conjugate base of the acid will be (consider molarity ≃ molality):

The correct option is C 9×10−11ΔTf=Kf×m=86×0.01=0.0186 i=0.02050.0186=1.10=1+α α=0.1 Ka=Cα21−α=19×10−3 Kb=9×10−11

$0.01 ~ M$ solution of an acid HA freezes at $- {0.0205}^{\circ} C$ . If $K_{f}$ for water is $1.86 {Kkgmol}^{- 1}$ .The ionization constant for the conjugate base of the acid will be (consider molarity $≃$ molality):

The correct option is C $9 \times 10^{- 11}$
$Δ T_{f} = K_{f} \times m = 86 \times 0.01 = 0.0186$ $i = \frac{0.0205}{0.0186} = 1.10 = 1 + α$
$α = 0.1$
$K_{a} = \frac{C α^{2}}{1 - α} = \frac{1}{9} \times 10^{- 3}$ $K_{b} = 9 \times 10^{- 11}$