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The solubility of Pb(OH)in water is 6.7 \times 10-6 M. Calculate the solubility of Pb(OH)2 in buffer solution of pH = 8.

Option: 1

1.2 \times 10^{-3} \mathrm{M}


Option: 2

1.2 \times 10^{+3} \mathrm{M}


Option: 3

2.4 \times 10^{-3} \mathrm{M}


Option: 4

1.2 \times 10^{-15}\mathrm{M}


Answers (1)

best_answer

\mathrm{Pb}(\mathrm{OH})_{2}=\mathrm{Pb}^{2+}+2 \mathrm{OH}^{-}\begin{array}{l}{\mathrm{Ksp}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2}} \\ {=6.7 \times 10^{-6} \times\left(2 \times 6.7 \times 10^{-6}\right)^{2}} \\ {=4 \times\left(6.7 \times 10^{-6}\right)^{3}} \\ {=1.2 \times 10^{-15}} \\ {\text { Solubility of } \mathrm{Pb}(\mathrm{OH})_{2} \text { in buffer system of } \mathrm{pH=}\: 8}\\ {\text { pH of buffer solution }=8} \\ {\mathrm{\: pOH}=6} \\ {\left[\mathrm{OH}^{-}\right]=10^{-6}} \\ {\left[\mathrm{Pb}^{2+}\right]=\frac{1.2 \times 10^{-15}}{\left(10^{-6}\right)_{2}}=1.2 \times 10^{-3} \mathrm{M}}\end{array}

Hence, option number (1) is correct.
 

Posted by

Shailly goel

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