Question

Which will make a basic buffer?
A) ${\text{50 ml of 0}}{\text{.1M NaOH + 25 ml of 0}}{\text{.1M C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$
B) ${\text{100ml of 0}}{\text{.1M C}}{{\text{H}}_{\text{3}}}{\text{COOH}} + 100{\text{ ml of 0}}{\text{.1M NaOH }}$
C) ${\text{100ml of 0}}{\text{.1M HCl}} + 200{\text{ ml of 0}}{\text{.1M N}}{{\text{H}}_{\text{4}}}{\text{OH }}$
D) ${\text{100ml of 0}}{\text{.1M HCl}} + 100{\text{ ml of 0}}{\text{.1M NaOH }}$

Answer 1

A solution that resists change in ${\text{pH}}$value upon the addition of a small amount of a strong acid or strong base is called a buffer solution. There are two types of buffer solution, acidic buffer solution and basic buffer solution. Write the acid- base reaction for all given mixtures. Using the volume and concentration of acids and bases determine the moles of reactant remain and moles of product from after the reaction. Using the species that remain in the reaction in the solution determine which solution will make a basic buffer solution.

Complete solution:
As we know that there are two types of buffer solution.
Acidic buffer: An acidic buffer solution consists of a solution of a weak acid and its salt with a strong base.
Basic buffer: A basic buffer solution consists of a solution of a weak base and its salt with strong acid.
So, out of the given mixtures we have to determine which mixture contains a weak base and salt with strong acid after the reaction.
Let us consider the mixture ${\text{50 ml of 0}}{\text{.1M NaOH + 25 ml of 0}}{\text{.1M C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$
The reaction between ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ and ${\text{NaOH }}$ is as follows:
${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH }} + {\text{ NaOH }} \rightleftharpoons {\text{ C}}{{\text{H}}_{\text{3}}}{\text{COONa + }}{{\text{H}}_{\text{2}}}{\text{O }}$
Now, we will calculate the initial moles of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ and ${\text{NaOH }}$.
${\text{moles = molarity }} \times {\text{ ml}}$
${\text{moles of C}}{{\text{H}}_{\text{3}}}{\text{COOH }} = 0.1{\text{ M }} \times {\text{25ml = 2}}{\text{.5mol}}$
${\text{moles of NaOH }} = 0.1{\text{ M }} \times 50{\text{ml = 5}}{\text{.0mol}}$
From the reaction, we can say that the mole ratio of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ and ${\text{NaOH }}$is 1:1. So 2.5 mol of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ reacts with 2.5 mol of ${\text{NaOH }}$and give 2.5mmol of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COONa}}$.
So, after the reaction solution contains 2.5 mmol of unreacted ${\text{NaOH }}$, 0 mol of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$and 2.5mmol of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COONa}}$.
So we can say that solution is basic due to the presence of an unreacted strong base ${\text{NaOH }}$but it cannot be a buffer solution as the solution does not contain${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$.
Hence, option (1) is an incorrect answer.
${\text{100ml of 0}}{\text{.1M C}}{{\text{H}}_{\text{3}}}{\text{COOH}} + 100{\text{ ml of 0}}{\text{.1M NaOH }}$
Now, we will calculate the initial moles of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ and ${\text{NaOH }}$.
${\text{moles of C}}{{\text{H}}_{\text{3}}}{\text{COOH }} = 0.1{\text{ M }} \times 100{\text{ml = 10mol}}$
${\text{moles of NaOH }} = 0.1{\text{ M }} \times 100{\text{ml = 10mol}}$
From the reaction, we can say that the mole ratio of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ and ${\text{NaOH }}$is 1:1. So 10 mol of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ reacts with 10 mmol of ${\text{NaOH }}$and give 10mol of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COONa}}$.
So, after the reaction solution contains only 10 mmol of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COONa}}$.
So, it cannot be a buffer solution as the solution does not contain${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$.
Hence, option (2) is an incorrect answer.
${\text{100ml of 0}}{\text{.1M HCl}} + 200{\text{ ml of 0}}{\text{.1M N}}{{\text{H}}_{\text{4}}}{\text{OH }}$
Now, we will calculate the initial moles of ${\text{HCl}}$ and ${\text{N}}{{\text{H}}_{\text{4}}}{\text{OH }}$.
${\text{moles of HCl }} = 0.1{\text{ M }} \times 100{\text{ml = 10mol}}$
${\text{moles of N}}{{\text{H}}_{\text{4}}}{\text{OH }} = 0.1{\text{ M }} \times 200{\text{ml = 20mmol}}$
From the reaction, we can say that the mole ratio of ${\text{HCl}}$ and ${\text{N}}{{\text{H}}_{\text{4}}}{\text{OH }}$is 1:1. So 10 mmol of ${\text{HCl}}$ reacts with 10 mmol of ${\text{N}}{{\text{H}}_{\text{4}}}{\text{OH }}$and give 10mmol of ${\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}$.
So, after the reaction solution contains only 10 mmol of unreacted${\text{N}}{{\text{H}}_{\text{4}}}{\text{OH }}$and 10mmol of ${\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}$.
As a mixture of ${\text{N}}{{\text{H}}_{\text{4}}}{\text{OH }}$and ${\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}$ form a basic buffer system we can say that mixture of
${\text{100ml of 0}}{\text{.1M HCl}} + 200{\text{ ml of 0}}{\text{.1M N}}{{\text{H}}_{\text{4}}}{\text{OH }}$will makes basic buffer solution.
Hence the correct option is (3) ${\text{100ml of 0}}{\text{.1M HCl}} + 200{\text{ ml of 0}}{\text{.1M N}}{{\text{H}}_{\text{4}}}{\text{OH }}$.
${\text{100ml of 0}}{\text{.1M HCl}} + 100{\text{ ml of 0}}{\text{.1M NaOH }}$
This is a mixture of strong acid and strong base.
Now, we will calculate the initial moles of ${\text{HCl}}$ and ${\text{NaOH }}$.
${\text{moles of HCl }} = 0.1{\text{ M }} \times 100{\text{ml = 10mmol}}$
${\text{moles of NaOH }} = 0.1{\text{ M }} \times 100{\text{ml = 10mmol}}$
From the reaction, we can say that the mole ratio of ${\text{HCl}}$ and ${\text{NaOH }}$is 1:1. So 10 mmol of ${\text{HCl}}$ react with 10 mmol of ${\text{NaOH }}$and give 10mmol of ${\text{NaCl}}$.
This is a neutralisation reaction.

Thus, the correct option is (C).

Note: A mixture of acid and base which gives a mixture of weak base and salt of strong acid after the reaction is known as a basic buffer. The addition of strong acid or strong base does not alter the ${\text{pH}}$ of the solution. The ${\text{pH}}$of buffer solution also remains unaffected after dilution.

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