Calculate the volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution. (Ksp of PbCl2 = 3.2 x 10-8, atomic mass of Pb = 207 u).

Answers (1)

As per the information available from the question, Ksp of PbCl2 = 3.2 ×10-8

Hence, we can write the equation of disassociation of PbCl2 as follows: -

PbCl_{2}\rightleftharpoons Pb^{+2}(aq)+2Cl^{-}(aq)

At t=0 1 0 0
At t=t 1-x x 2x

K_{sp}=[Pb^{2+}][Cl^{-}]^{2}

On substituting the given values, we get

=(x)\times (2x)^{2}=4x^{3}

4x^{3}=3.2\times 10^{-8}

x=2\times 10^{-3}mol/L

We know that solubility is a product of molar mass (PbCl_{2}) \times 2 \times 10^{-3}

=556\times 10^{-3}=0.556g/L

0.1g of PbCl_{2} will be dissolved in 0.1798 L (0.1/0.0556)

Thus, 0.1798 is the required volume to get a saturated solution of PbCl_{2}

 


 

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