If {K}_{ {sp}} ( {PbSO}_{4} )=1.8 \times 10^{-8} \& {K}_{ {a}} ( {HSO}_{4}^{⊖} )= 1\times10^{-2} the equilibrium constant for the reaction PbS_4(s)+H^{⊖}(aq)⇌HSO^{⊕}_4(aq)+Pb^{2+}(aq)

Correct Answer is : 1.8×10−6

NEET 2019 Model Question Paper 4

Question : 81 of 180

Marks: +1, -0

If Ksp(PbSO4)=1.8×10−8&‌Ka(HSO4⊖)=1×10−2 the equilibrium constant for the reaction
PbS4(s)+H⊖(aq)⇌HS

⊕

4(aq)+Pb2+(aq)

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