Question

$\text{50 mL }$ of $\text{0}\text{.2 M }$ ammonia solution is treated with $\text{25 mL }$of$\text{0}\text{.2 M HCl }$. If $\text{p}{{\text{K}}_{\text{b}}}$ of ammonia solution is 4.75, the$\text{ pH }$ of the mixture will be:
(A) 3.75
(B) 9.25
(C) 8.25
(D) 4.75

Answer 1

To solve this question, we need to have an idea about the number of moles present in ammonia and hydrochloric acid. Once we know the number of moles, we can now find the concentration. For a mixture of solution, the concentration of conjugate acid and weak base is always equal.

Complete step by step answer:
Let us first write the equation of the reaction that is mentioned in the question. So it is:
$\text{HCl (aq) + N}{{\text{H}}_{\text{3}}}\text{ (aq) }\xrightarrow{{}}\text{ N}{{\text{H}}_{\text{4}}}^{\text{+}}\text{ (aq) + C}{{\text{l}}^{-}}\text{ (aq)}$

Let us calculate the number of moles of ammonia and hydrochloric acid.
The number of moles of $\text{N}{{\text{H}}_{\text{3}}}$ is,
$\text{No}\text{. of moles of N}{{\text{H}}_{\text{3}}}\text{ = 50 }\\!\\!\times\\!\\!\text{ }\dfrac{\text{0}\text{.2}}{\text{1000}}\text{ = 0}\text{.01 mol}$
The number of moles of $\text{HCl }$ is,
$\text{No}\text{.of moles of HCl = 25 }\\!\\!\times\\!\\!\text{ }\dfrac{\text{0}\text{.2}}{\text{1000}}\text{ = 0}\text{.005 mol }$

Here, the moles of $\text{HCl }$ is less than that of the moles of $\text{N}{{\text{H}}_{\text{3}}}$ therefore $\text{HCl }$ is a limiting reagent. The amount of $\text{HCl }$ decides the amount of product formed.
Then the number of moles of $\text{N}{{\text{H}}_{\text{3}}}$ reacting completely is equal to,$\text{No}\text{.of moles of N}{{\text{H}}_{\text{3}}}\text{ reacted completely = 0}\text{.01 - 0}\text{.005 = 0}\text{.005 mol}$
So the total volume of the solution will be equal to the sum of the volume of ammonia and$\text{HCl }$.
$\text{Total volume (V) = }50\text{ + }25\text{ }=\text{ 75 ml}$
So now we have to find the concentration. We know that,
$\text{Concentration = }\dfrac{\text{n}}{\text{V}}=\dfrac{\text{0}\text{.005}\times 1000\text{ mL}}{75\text{ L}}\text{ = 0}\text{.07 mol }{{\text{L}}^{\text{-1}}}$
Now we have to find the value of $\text{ p}{{\text{K}}_{\text{b}}}$.

& \text{p}{{\text{K}}_{\text{b}}}\text{ = -log (}{{\text{K}}_{\text{b}}}) \\\ & \text{4}\text{.75 = log (}{{\text{K}}_{\text{b}}}) \\\ & {{10}^{-4.75}}=\text{ }{{\text{K}}_{\text{b}}} \\\ & \therefore {{\text{K}}_{\text{b}}}\text{ =1}\text{.8 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{-\text{5}}} \\\ \end{aligned}$$ Now we know that Henderson –Hasselbach equation given as, $\text{pOH = p}{{\text{K}}_{\text{b}}}\text{ + log }\dfrac{\left[ \text{conjugate acid} \right]}{\left[ \text{weak base} \right]}\text{ }$ But keeping in mind the question, we have to consider the fact that there are equal concentrations of conjugate acid and weak base. Since the concentrations of conjugate acid and weak base are equal, The Henderson equation becomes, $\text{pOH = p}{{\text{K}}_{\text{b}}}\text{ }$ We have considered the amount to be 0 while performing the calculations. Now we have to find the value of$$\text{ pOH}$$. Let us substitute the values to get the value of$$\text{ pOH }$$. $\text{pOH = - log 1}\text{.8 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{-\text{5}}}\text{ = p}{{\text{K}}_{\text{b}}}\text{= 4}\text{.74}$ So the value of pH is, $\text{ pH = 14}-\text{pOH = 14 - 4}\text{.74 = 9}\text{.25 }$ Therefore, the value of $\text{pH}$ is 9.25. **So, the correct answer is “Option B”.** **Note:** We should be having an idea about the concept of conjugate acid and a weak base. By conjugate acid, we mean a compound which is formed by the reception of a proton or as we know ${{\text{H}}^{\text{+}}}$ by a base. A weak base is defined to be a chemical base that does not ionize completely in an aqueous solution.