Question

Which will make the basic buffer?
A) ${\text{50 mL of 0}} \cdot {\text{1 M NaOH + 25 mL of 0}} \cdot {\text{1M C}}{{\text{H}}_3}COOH$
B) ${\text{100 mL of 0}} \cdot {\text{1 M C}}{{\text{H}}_3}{\text{COOH + 100 mL of 0}} \cdot {\text{1M NaOH}}$
C) ${\text{100 mL of 0}} \cdot {\text{1 M HCl + 200 mL of 0}} \cdot {\text{1M N}}{{\text{H}}_4}OH$
D) ${\text{100 mL of 0}} \cdot {\text{1 M HCl + 100 mL of 0}} \cdot {\text{1M NaOH}}$

Answer 1

Hint:A basic buffer solution is a solution made up of a weak base and that weak base’s salt. The volume and molarity of each reacting component are given which means one can calculate the number of moles that reacted and moles formed of products. One can compare them and find out the weak base and its salt and make the correct choice.

Complete step by step answer:
1) First of all we will learn about the concept of the basic buffer solution. The basic buffer solution can be elaborated as the solution which contains the weak base and its salt. Now, let us analyze the given options and find out the moles of reactants and products in each option.
2) In option A the reaction will be as,
$NaOH + C{H_3}COOH \to C{H_3}COONa + {H_2}O$
Before the reaction, it will have the moles as $NaOH = 50 \times 0 \cdot 1 = 5mmol$ and $C{H_3}COOH = 25 \times 0 \cdot 1 = 2 \cdot 5mmol$. After the reaction, there will remain only ${\text{2}} \cdot {\text{5mmol}}$ of ${\text{NaOH}}$ when an equal amount of reactants gets reacted and excess remains. The product formed will be $C{H_3}COONa = 2 \cdot 5mmol$.
The above reaction solution will be basic due to the sodium hydroxide base and it is not a basic buffer solution.
3) In option B ${\text{100 mL of 0}} \cdot {\text{1 M C}}{{\text{H}}_3}{\text{COOH + 100 mL of 0}} \cdot {\text{1M NaOH}}$ the equal amounts of both the reactants will form an equal amount of the product $C{H_3}COONa = 10mmol$. As the solution will have the same amounts of acid, base, and salt this solution is not a buffer solution.
4) In option C ${\text{100 mL of 0}} \cdot {\text{1 M HCl + 200 mL of 0}} \cdot {\text{1M N}}{{\text{H}}_4}OH$ the products formed will be $N{H_4}Cl = 10mmol$ and the more amount of base ammonium hydroxide with its salt makes the solution as a basic buffer. Hence, this option is the correct choice.
5) In option D ${\text{100 mL of 0}} \cdot {\text{1 M HCl + 100 mL of 0}} \cdot {\text{1M NaOH}}$ the product salt formed will be in equal amount with the reactants of acid and base and hence, it is not a basic buffer solution.
Therefore, the basic buffer will be of ${\text{100 mL of 0}} \cdot {\text{1 M HCl + 200 mL of 0}} \cdot {\text{1M N}}{{\text{H}}_4}OH$ which shows option C as the correct choice.

Note:
If one analyzes all the options carefully, there is sodium hydroxide based on three options which is a strong base. As we know basic buffers can only be made by the weak base which is ammonium hydroxide given in option C that is the correct choice. This identification of only a weak base in the options method can be less time-consuming however one must analyze the options carefully.