Calculate the molar solubility of Ni(OH) 2 in 0·10M NaOH. The ionic product of Ni(OH) 2 is 2·0 × 10 –15 . - Wired Faculty

Question

Calculate the molar solubility of Ni(OH)₂ in 0·10M NaOH. The ionic product of Ni(OH)₂ is 2·0 × 10^–15.

Name: Wired Faculty - The Best Learning App
Rating: 4.6 (1864 reviews)

Solution

Let the solubility of Ni(OH)₂ in 0·10M NaOH = S mol L^–1.

Ni left parenthesis OH right parenthesis subscript 2 space space space space space space space rightwards harpoon over leftwards harpoon space space space space space Ni to the power of 2 plus end exponent space space space space space plus space space space space space space space space space 2 OH to the power of minus straight S space space space space space space space space space space space space space space space space space space space space space space space space space straight S space space space space space space space space space space space space space space space space space space space space space left parenthesis 0.10 space plus space 2 straight S right parenthesis

Applying the law of solubility product

straight K subscript sp space equals space open square brackets Ni to the power of 2 plus end exponent close square brackets space open square brackets OH to the power of minus close square brackets squared

straight K subscript sp space equals space straight S space cross times space left parenthesis 0.10 space plus space 2 straight S right parenthesis squared

straight i. straight e. space 2.0 space cross times space 10 to the power of negative 15 end exponent space equals space straight S space cross times space left parenthesis 0.10 right parenthesis squared space left square bracket As space 2 straight S less than 0.10 right square bracket or space space space space space straight S space equals fraction numerator 2.0 space cross times space 10 to the power of negative 15 end exponent over denominator left parenthesis 0.10 right parenthesis squared end fraction space space space space space space space space space space equals 2.0 space cross times space 10 to the power of negative 13 end exponent mol space straight L to the power of negative 1 end exponent space space space space space space space space space space equals 2.0 space cross times space 10 to the power of negative 13 end exponent straight M

For Daily Free Study Material Join wiredfaculty