
1 M $\text{N}{{\text{H}}_{\text{4}}}\text{OH}$ and 1 M $\text{HCl}$ are mixed to make a total volume of 300 mL. If the pH of the mixture is $9.26$ and $\text{p}{{\text{K}}_{\text{a}}}$ of $\text{N}{{\text{H}}_{\text{4}}}^{\text{+}}$ = $9.26$ then the volume ratio of $\text{N}{{\text{H}}_{\text{4}}}\text{OH}$and$\text{HCl}$ will be:
A.1:1
B.1:2
C.2:1
D.3:1
Answer
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Hint:The Henderson-Hasselbalch equation is used to determine the pH and the pOH of different buffer solutions and this equation is commonly used in different biochemistry and chemistry equations. The equation can be mathematically represented as:
Formula used:
$\text{pH = p}{{\text{K}}_{\text{a}}}\text{ + log}\left[ \dfrac{\text{salt}}{\text{acid}} \right]$, where pH is the measure of the hydrogen ions in the solution.
Complete step by step answer:
In the above solution, on mixing 1 M$\text{N}{{\text{H}}_{\text{4}}}\text{OH}$ and 1 M $\text{HCl}$ react together to form ammonium chloride,$\text{N}{{\text{H}}_{\text{4}}}\text{Cl}$ and along with the remaining ammonium acid in the medium that does not react with the hydrochloric acid, it forms the buffer solution. This is because ammonium hydroxide is a weak base while hydrochloric acid is a strong acid. Therefore the pH of the buffer solution is,
$\text{9}\text{.26 = 9}\text{.26 + log}\left[ \dfrac{\text{salt}}{\text{acid}} \right]$
Or, $\text{log}\left[ \dfrac{\text{salt}}{\text{acid}} \right]=0$
Or, $\text{log}\left[ \dfrac{\text{salt}}{\text{acid}} \right]\text{= log }\left[ \text{1} \right]$
Cross-multiplying:
Or, $\left[ \text{salt} \right]\text{= }\left[ \text{acid} \right]$
As one mole of hydrochloric acid neutralizes one mole of ammonium hydroxide to form one mole of ammonium chloride, therefore the solution must have 200 mL of ammonium hydroxide and 100 mL of hydrochloric chloride so that 100 mL of ammonium chloride could be formed.
Thus, the volume ratio of $\text{N}{{\text{H}}_{\text{4}}}\text{OH}$ and $\text{HCl}$ will be 2:1, option C.
Note:
In 1908, Lawrence Joseph Henderson derived an equation to calculate the pH of the buffer solution. In 1917, Karl Albert Hasselbalch re-expressed that same formula in the terms of a logarithmic equation resulting in the Henderson –Hasselbalch equation. This equation can be used to calculate the pH of the buffer consisting of the conjugate acid and its base or conjugate base and its acid.
Formula used:
$\text{pH = p}{{\text{K}}_{\text{a}}}\text{ + log}\left[ \dfrac{\text{salt}}{\text{acid}} \right]$, where pH is the measure of the hydrogen ions in the solution.
Complete step by step answer:
In the above solution, on mixing 1 M$\text{N}{{\text{H}}_{\text{4}}}\text{OH}$ and 1 M $\text{HCl}$ react together to form ammonium chloride,$\text{N}{{\text{H}}_{\text{4}}}\text{Cl}$ and along with the remaining ammonium acid in the medium that does not react with the hydrochloric acid, it forms the buffer solution. This is because ammonium hydroxide is a weak base while hydrochloric acid is a strong acid. Therefore the pH of the buffer solution is,
$\text{9}\text{.26 = 9}\text{.26 + log}\left[ \dfrac{\text{salt}}{\text{acid}} \right]$
Or, $\text{log}\left[ \dfrac{\text{salt}}{\text{acid}} \right]=0$
Or, $\text{log}\left[ \dfrac{\text{salt}}{\text{acid}} \right]\text{= log }\left[ \text{1} \right]$
Cross-multiplying:
Or, $\left[ \text{salt} \right]\text{= }\left[ \text{acid} \right]$
As one mole of hydrochloric acid neutralizes one mole of ammonium hydroxide to form one mole of ammonium chloride, therefore the solution must have 200 mL of ammonium hydroxide and 100 mL of hydrochloric chloride so that 100 mL of ammonium chloride could be formed.
Thus, the volume ratio of $\text{N}{{\text{H}}_{\text{4}}}\text{OH}$ and $\text{HCl}$ will be 2:1, option C.
Note:
In 1908, Lawrence Joseph Henderson derived an equation to calculate the pH of the buffer solution. In 1917, Karl Albert Hasselbalch re-expressed that same formula in the terms of a logarithmic equation resulting in the Henderson –Hasselbalch equation. This equation can be used to calculate the pH of the buffer consisting of the conjugate acid and its base or conjugate base and its acid.
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