Question

If the maximum concentration of pbcl2 in water is 0.01 m at 298 k , its maximum concentration in 0.1 m nacl will be

Answer 1

Answer : The maximum concentration of PbCl₂ in 0.1 M NaCl is 0.0004 M

Explanation:

Step 1 : Find concentrations of the ions

The dissociation reaction for PbCl2 in water can be write as,

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

The maximum concentration of PbCl2 in water is 0.01 M

We can find concentration of the ions using the stoichiometry from the above reaction

The concentration of Pb²⁺ ion is 0.01 M and concentration of Cl⁻ ion is 2 x 0.01 = 0.02 M.

Step 2 : Find solubility product

The solubility product of PbCl₂ can be written as,

$Ksp = [Pb^{2+}][Cl^{-}]^{2}$

Let us plug in the concentrations of Pb and Cl.

$Ksp = (0.01)(0.02)2$

Ksp = 4.0 x 10⁻⁶

Step 3 : Find new concentration of Pb²⁺ using common ion effect.

When PbCl₂ is dissolved in 1 M NaCl, there is a common ion Cl-. This decreases the solubility of PbCl₂ due to the common ion effect.

The concentration of Cl- in 0.1 M NaCl is 0.1. Let us plug in this value to find the new concentration of Pb²⁺ ion.

$Ksp = [Pb^{2+}][Cl^{-}]^{2}$

$4 \times 10^{-6}= [Pb^{2+}](0.1)^{2}$

$[Pb^{2+}]= \frac{4\times 10^{-6}}{0.01}$

$[Pb^{2+}]= 0.0004 M$

Pb²⁺ ion comes from dissolved PbCl₂. Therefore the concentration of PbCl₂ is same as that of Pb²⁺

The maximum concentration of PbCl₂ in 0.1 M NaCl is 0.0004 M

Answer 2

Answer:

4×10-4 M

HOPE THIS HELPS YOU:-)