Question

Zns solubility in water 0.97g/l we mixed 2 moles of zn(N03)2 in 5L solution find new solubility of Zns

Answer 1

To find the Solubility of Zns

Further Explanation:

Given:

ZnS solubility in water = 0.97g/l

= $\frac{0.97}{97}$ mol/L         [Molar weight of ZnS = 97 moles/g]

= 0.01 M

Now,

Solubility product Ksp = (0.01 M) × (0.01 M)

= 10⁻⁴ M²

now,

If we mix 2 moles of Zn(NO3)2

i.e 2 moles of Zn²⁺ ions in 5L solution,

then the added concentration of Zn²⁺ ions will be

= $\frac{2}{5}$

= 0.4 M

$ZnS(s)\ \rightleftharpoons\ Zn^{2+}(aq)\ +\ S^{2-}(aq)$

Conc.         s + 0.4                   s              Ksp = (s + 0.4)(s)

$\Rightarrow 10{-4} =(s + 0.4)(s)\Rightarrow\ s^2+(0.4) s-10^{-4}=0$

$s=\frac{[- 0.4 + \sqrt{(0.4)^2 + 4\times10^-^4}]}{2} =\frac{[-0.4 + \sqrt{16\times10^-^2 + 4\times10^-^4}]}{2}$

$=\frac{[-0.4 + 10^-^2 \sqrt{1604}]}{2}= -0.2+ 10^-^2\sqrt{401}$

 = -0.2 + 0.20025  

=  0.00025

s = 2.5 × 10⁻⁴ M

or

= 2.5 × 97 × 10⁻⁴ g/L

= 0.02425 g/L

Hence,

the new solubility of ZnS is 0.02425 g/L

Learn more: Find the solubility

The slightly soluble ionic compound PbF2 has a solubility in water of 0.74 g/L. What is the numerical value of Ksp for PbF2?  https://brainly.com/question/13700468

Answer 2

Answer:

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