Question

At 25 oC the solubility of mercury(II) sulfide is 1.26 x 10-26 mol/L. Calculate the value of Ksp at this temperature. Give your answer in scientific notation to 2 SIGNIFICANT FIGURES (even though this is strictly incorrect).

Answer 1

Answer: The solubility product of mercury (II) sulfide is $1.6\times 10^{-52}$

Explanation:

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The chemical equation for the ionization of mercury (II) sulfide follows:

$HgS(aq.)\rightleftharpoons Hg^{2+}(aq.)+S^{2-}(aq.)$  

                      s                  s

The expression of $K_{sp}$ for above equation follows:

$K_{sp}=s\times s$

We are given:  

$s=1.26\times 10^{-26}M$

Putting values in above expression, we get:

$K_{sp}=(1.26\times 10^{-26})\times (1.26\times 10^{-26})\\\\K_{sp}=1.59\times 10^{-52}$

Hence, the solubility product of mercury (II) sulfide is $1.6\times 10^{-52}$

Answer 2

TThe value of Ksp at 25 degrees celius temperature is mathematically given as

Ksp=1.59* 10^{-52}

What is the value of Ksp at 25 degrees Celius temperature.?

Question Parameter(s):

At 25 oC the solubility of mercury(II) sulfide is 1.26 x 10-26 mol/L.

Generally, the equation for the Chemical Reaction  is mathematically given as

HgS(aq).------>  Hg^{2+}(aq.)+S^{2-}(aq.)  

Where

Ksp=s* s

Therefore

Ksp=(1.26*10^{-26})*(1.26\times 10^{-26})

Ksp=1.59* 10^{-52}

In conclusion, solubility is

Ksp=1.59* 10^{-52}

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