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Question: How many gm of solid KOH must be added to 100 ml of a buffer solution, which is 0.1 M each with resp...

How many gm of solid KOH must be added to 100 ml of a buffer solution, which is 0.1 M each with respect to acid H A and salt K A to make the pH of the solution 6.0?
[Given pKa(HA)p{{K}_{a}}\left( HA \right) = 5]

Explanation

Solution

The buffer solution resists the changes in the concentration of hydrogen ions when we add acid or base into it (here KOH). Thus, maintaining the pH value.

Complete step by step solution:
As we know, that the buffer solution maintains the constant pH thus, they either possess reserved acidity or reserved basicity. Buffer solutions are of two types i.e.
1. Acidic buffer- contains equimolar quantities of weak acid and its salt along with strong base. The pH of acidic buffer can be determined by,
pH=pKa+log⁡[salt][acid]pH=p{{K}_{a}}+\log \dfrac{\left[ salt \right]}{\left[ acid \right]}
where, pKap{{K}_{a}}= acidic constant of acid.
2. Basic buffer- contains equimolar quantities of weak base and its salt along with strong acid. The pH of basic buffer can be determined by,
pH=14−pKb−log⁡[salt][base]pH=14-p{{K}_{b}}-\log \dfrac{\left[ salt \right]}{\left[ base \right]}
where, pKbp{{K}_{b}} = basic constant of base.
Now, focusing on the illustration we get,
Given that,
Molarity of buffer solution = 0.1 M
Volume of buffer solution = 100 ml
pH = 6
pKa(HA)p{{K}_{a}}\left( HA \right) = 5
As KOH is added to the buffer, it results in the formation of an acidic buffer.
Thus,
HA+OH−→A−+H2OHA+O{{H}^{-}}\to {{A}^{-}}+{{H}_{2}}O
By the stoichiometry of the above equation,
Number of moles of HA = number of moles of A−{{A}^{-}}= 0.1M×100ml1000\dfrac{0.1M\times 100ml}{1000} = 0.01 moles
Now,
HA+OH−→A−+H2OHA+O{{H}^{-}}\to {{A}^{-}}+{{H}_{2}}O
0.01 0.001 ______ initially
(0.001-x) (0.001+x) ______ when KOH is added simultaneously.
Thus, using equation for acidic buffer,
pH=pKa+log⁡[salt][acid]pH=p{{K}_{a}}+\log \dfrac{\left[ salt \right]}{\left[ acid \right]}
⇒pH=pKa+log⁡[A−][HA] ⇒pH=pKa+log⁡(0.001+x)(0.001−x) ⇒6=5+log⁡(0.001+x)(0.001−x) ⇒(0.001+x)(0.001−x)=antilog⁡1=10  \begin{aligned} & \Rightarrow pH=p{{K}_{a}}+\log \dfrac{\left[ {{A}^{-}} \right]}{\left[ HA \right]} \\\ & \Rightarrow pH=p{{K}_{a}}+\log \dfrac{\left( 0.001+x \right)}{\left( 0.001-x \right)} \\\ & \Rightarrow 6=5+\log \dfrac{\left( 0.001+x \right)}{\left( 0.001-x \right)} \\\ & \Rightarrow \dfrac{\left( 0.001+x \right)}{\left( 0.001-x \right)}=anti\log 1=10 \\\ & \\\ \end{aligned}
Thus, x = 8.18 moles of KOH
Molar mass of KOH = 56 gm/mol.

Therefore, weight of KOH added = 8.18×56=458Kg=0.458gm8.18\times 56=458Kg=0.458gm

Note: Make a note that the pH of the buffer solution is not affected by the dilution as the ratio under logarithm remains the same.