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Question: Calculate the degree of hydrolysis of the 0.01M solution of salt \(KF\) (\({K_{HF}} = 6.6 \times {10...

Calculate the degree of hydrolysis of the 0.01M solution of salt KFKF (KHF=6.6×10−4{K_{HF}} = 6.6 \times {10^{ - 4}}):
A. 3.87×10−63.87 \times {10^{ - 6}}
B. 3.87×10−53.87 \times {10^{ - 5}}
C. 3.87×10−23.87 \times {10^{ - 2}}
D. None of these

Explanation

Solution

Hydrolysis is a process in which compound reacts with water to decompose and this results in the production of acidity or alkalinity. The degree of hydrolysis of salt is defined as the fraction of the total salt which is hydrolyzed. Mathematically it is given as:
h=No. of moles of salt hydrolysistotal number of moles of salth = \dfrac{{{\text{No}}{\text{. of moles of salt hydrolysis}}}}{{{\text{total number of moles of salt}}}}

Complete step by step answer:
As per the question hydrolysis constant (KHF{K_{HF}}) is given as 6.6×10−46.6 \times {10^{ - 4}}and the hydrolysis reaction of KFKF is given below:
KF+H2O→HF+KOHKF + {H_2}O \to HF + KOH
Initially i.e. at t=0
0.01M 0 0\text{0.01M 0 0}
At time t
0.01(1-h) 0.01h 0.01\text{0.01(1-h) 0.01h 0.01}
Here ‘h’ is the degree of hydrolysis. The concentration of HF and KOH at time ‘t’ is calculated as:
h=No. of moles of salt hydrolysistotal number of moles of salth = \dfrac{{{\text{No}}{\text{. of moles of salt hydrolysis}}}}{{{\text{total number of moles of salt}}}}
⇒h=No. of moles of salt hydrolysis0.01\Rightarrow h = \dfrac{{{\text{No}}{\text{. of moles of salt hydrolysis}}}}{{0.01}}
⇒0.01×h=No. of moles of salt hydrolysis\Rightarrow 0.01 \times h = {\text{No}}{\text{. of moles of salt hydrolysis}}
Or
No. of moles of salt hydrolysis{\text{No}}{\text{. of moles of salt hydrolysis}}=0.01h
Since the hydrolysis constant can be found as:
KHF=[HF]×[KOH][KF]{K_{HF}} = \dfrac{{[HF] \times [KOH]}}{{[KF]}}
On putting the value of the molar concentration of KF, HF, and KOH in the above equation we get:
KHF=0.01h×0.01h0.01(1−h){K_{HF}} = \dfrac{{0.01h \times 0.01h}}{{0.01(1 - h)}}
⇒6.6×10−4=0.01h×0.01h0.01(1−h)\Rightarrow 6.6 \times {10^{ - 4}} = \dfrac{{0.01h \times 0.01h}}{{0.01(1 - h)}}
Since the value of h is very small so (1-h) is considered to be equal to 1.
⇒6.6×10−4=0.01h×0.01h0.01\Rightarrow 6.6 \times {10^{ - 4}} = \dfrac{{0.01h \times 0.01h}}{{0.01}}
⇒6.6×10−4=0.0001h20.01\Rightarrow 6.6 \times {10^{ - 4}} = \dfrac{{0.0001{h^2}}}{{0.01}}
⇒6.6×10−4×0.010.0001=h2\Rightarrow \dfrac{{6.6 \times {{10}^{ - 4}} \times 0.01}}{{0.0001}} = {h^2}
⇒660×10−4=h2\Rightarrow 660 \times {10^{ - 4}} = {h^2}
⇒660×10−4=h\Rightarrow \sqrt {660 \times {{10}^{ - 4}}} = h
⇒h=25.69×10−2\Rightarrow h = 25.69 \times {10^{^{ - 2}}}
Thus, option D is the correct answer.

Additional information:
The degree of dilution does not affect the degree of hydrolysis of salts of weak acid and bases as there is mention of concentration in hydrolysis. Also, the degree of hydrolysis is directly proportional to the temperature as with an increase in temperature the overall increase of dissociation constant of water is much greater compared to the change in dissociation constant of acid or base.

Note: The salt of strong acid and the strong base does not undergo hydrolysis thus the talk of hydrolysis constant and degree of hydrolysis in the case of the salt of strong acid and strong base is meaningless.