Solubility products of four sparingly soluble salts {AB}, {A}_{2}, {A B}_{3} and {A}_{2} {B}_{3} are 1.0 \times10^{-10}, 4.0\times 10^{-15}, 2.7\times 10^{-19} and 1.08 \times10^{-3} respectively. The correct order of solubilities of these salts in water are

Correct Answer is : AB=A2B=AB3=A2B

{AB} > {A}_{2} {B} > {AB}_{3} > {A}_{2} {B}_{3}

{AB}= {A}_{2} {B}= {AB}_{3}= {A}_{2} {B}

{A}_{2} {B}_{3} > {AB}_{3 } > {A}_{3} {B} > {AB}

{AB} > {AB}_{3} > {A}_{2} {B} > {A}_{2} {B}_{3}

AP and TS EAMCET Medical Model Paper 6

Question : 142 of 160

Marks: +1, -0

Solubility products of four sparingly soluble salts AB,A2,AB3 and A2B3 are 1.0×10−10,4.0×10−15,2.7×10−19 and 1.08×10−3 respectively. The correct order of solubilities of these salts in water are

AB>A2B>AB3>A2B3
AB=A2B=AB3=A2B
A2B3>AB3>A3B>AB
AB>AB3>A2B>A2B3

Validate

Go to Question:

Prev Question

Next Question