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Question: In a solution \(0.04M\) \(FeC{l_3}\). \[0.02M\] \(FeC{l_3}\) and \(0.01M\) \(HCl\). How large may be...

In a solution 0.04M0.04M FeCl3FeC{l_3}. 0.02M0.02M FeCl3FeC{l_3} and 0.01M0.01M HClHCl. How large may be its pH{\text{pH}} (nearest integer) with out there precipitation of either FeSO4{\text{FeS}}{{\text{O}}_4} or FeSO4{\text{FeS}}{{\text{O}}_4}? Ksp{{\text{K}}_{{\text{sp}}}} of FeSO4{\text{FeS}}{{\text{O}}_4} and Fe(OH)3{\text{Fe}}{\left( {{\text{OH}}} \right)_3} are 8×10−10 and 4×10−328 \times {10^{ - 10}}{\text{ and 4}} \times {\text{1}}{{\text{0}}^{ - 32}} respectively.

Explanation

Solution

As we know that the acidity or basicity of a compound is determined by the concentration of H+ ions. The pHpH of the solution is calculated from the value of pOHpOH

Formula used: The formula to calculate the pHpH of the solution is,
pOH+pH=14pOH + pH = 14
pH=14−pOHpH = 14 - pOH

Complete step by step answer:
Given data contains,
Since the solubility product of ferric hydroxide has lower value it will precipitate first The expression for the solubility product is $$$$ Ksp{K_{sp}} [Fe3+][OH−]3.[F{e^{3 + }}]{[O{H^ - }]^3}.
Substitute values in the above expression.
4×10−38=0.02×[OH−]34 \times {10^{ - 38}} = 0.02 \times {[{\rm O}{{\rm H}^ - }]^3}
Hence [OH−]=1.25×10−12[O{H^ - }] = 1.25 \times {10^{ - 12}}
⇒pOH=−log⁡[OH−]\Rightarrow pOH = - \log [O{H^ - }]
When we substitute the value of [OH−][O{H^ - }] we get,
⇒pOH=−log⁡1.25×10−12\Rightarrow pOH = - \log 1.25 \times {10^{ - 12}}
⇒pOH=11.90\Rightarrow pOH = 11.90
Now apply the value pOH we get,
⇒pH=14−pOH\Rightarrow pH = 14 - pOH
⇒pH=14−11.90\Rightarrow pH = 14 - 11.90
⇒pH=2.10\Rightarrow pH = 2.10

Additional Information:
Now we discuss about the concept of solubility product as,
Solubility Product:
Solubility product is the equilibrium constant of a chemical reaction in which a solid ionic compound dissolves to yield its ions in solution. It is denoted by ‘Ksp{{\text{K}}_{{\text{sp}}}}’.
Solubility products are also known as “ion products”.
The value of solubility products usually increases with an increase in temperature due increased solubility.
Some of the factors that affect the value of solubility products are the common-ion effect, the diverse-ion effect and presence of ion-pairs.
If the ionic concentrations are lesser than the solubility product, the solution isn’t saturated. No precipitate would be formed.

Note:
We can define pH as the power of hydrogen ions in a given solution. We can calculate pH of the solution using a formula,
pH=−log⁡10[H+]pH = - \log 10\left[ {{H^ + }} \right]
If the pH value of 77 shows as neutral as water, pH of the solution is 1414 then the solution is highly basic, 00 means the solution is highly acidic. Acids are sour taste, it turns blue litmus to red, example vinegar (acetic acid).Bases are bitter taste, it turns red litmus to blue, example baking soda.