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Question: Calculate the amount of \(N{H_3}\) and \(N{H_4}Cl\) required to prepare a buffer Solution of \(pH\) ...

Calculate the amount of NH3N{H_3} and NH4ClN{H_4}Cl required to prepare a buffer Solution of pHpH 9.09.0 when total concentration of buffering reagents is 0.6molL10.6mol{L^{ - 1}}. pKbp{K_b} for NH3=4.7,log2=0.30N{H_3} = 4.7,\log 2 = 0.30

Explanation

Solution

pOH+pH=14pOH + pH = 14. Therefore, if we know the value of one, the other can be found out just by subtracting it from 14. Ammonium chloride is the salt of the reaction between ammonium hydroxide, a base, and hydrochloric acid, an acid. A buffer Solution, also commonly referred to as a pHpH buffer or hydrogen ion buffer, is an aqueous Solution consisting of a mixture of a weak acid and its conjugate base, or vice versa. The pHpH change in a buffer Solution is very negligible when a small amount of strong acid or base is added to it.

Formulae used: pOH=pKb+log[Salt][Base]pOH = p{K_b} + \log \dfrac{{[Salt]}}{{[Base]}}
Where pOHpOH is the negative log of the concentration of hydroxyl ions, pKbp{K_b} is the negative log of the base dissociation constant and [Salt],[Base][Salt],[Base] are the respective concentrations of the salt and the base present in the buffer Solution.

Complete step by step answer:
From the following equation, we have a relation between pOHpOH and pHpH:
pOH+pH=14pOH=14pHpOH + pH = 14 \Rightarrow pOH = 14 - pH
As we know that the buffer Solution should have a pHpH of 9, its pOHpOH value is:
pOH=149=5pOH = 14 - 9 = 5
Substituting this in our equation for pOHpOH, we get:
5=4.7+log[Salt][Base]5 = 4.7 + \log \dfrac{{\left[ {Salt} \right]}}{{[Base]}}
Simplifying the above result, we get:
log[Salt][Base]=54.7=0.3\log \dfrac{{[Salt]}}{{[Base]}} = 5 - 4.7 = 0.3
But we have already been given that 0.3=log20.3 = \log 2
[Salt][Base]=2[Salt]=2×[Base]\Rightarrow \dfrac{{[Salt]}}{{[Base]}} = 2 \Rightarrow [Salt] = 2 \times [Base] (1)
The total concentration of the buffering reagents, as given in the question is 0.60.6. Therefore,
[Salt]+[Base]=0.6[Salt] + [Base] = 0.6 (2)
Substituting equation 1 in equation 2, we get:
3×[Base]=0.6[Base]=0.23 \times [Base] = 0.6 \Rightarrow [Base] = 0.2
Substituting this value in equation 1, we get:
[Salt]=2×0.2=0.4[Salt] = 2 \times 0.2 = 0.4
Thus, the concentrations required are:
NH3=0.2mol/L NH4Cl=0.4mol/L  N{H_3} = 0.2mol/L \\\ N{H_4}Cl = 0.4mol/L \\\

Additional Information: Ka,pKa,Kb{K_a},p{K_a},{K_b}and pKbp{K_b} help us to predict whether a given species will accept or donate protons at the specified pHpH level. They give us information about the degree of dissociation of an acid or base. Ka{K_a} and pKap{K_a} relate to acids, while Kb{K_b} and pKbp{K_b} deal with bases. They give us relationships between the concentrations of hydrogen ions and hydroxyl ions, which are measurable quantities. Finding out these values can be of great importance in industries, where even a slight change in these quantities can lead to a drastic change in output.

Note: Buffer Solutions have the tendency to resist the changes in pHpH. Acidic buffers are made from weak acids, and a corresponding salt, while basic buffers, like in this question, are made from a weak base and its corresponding salt. Always remember that we have to take the negative logarithm. Like the pOHpOH equation, many others are there which describe various interdependencies. And it is important to identify in each question, which is the base and salt, so that the right concentration terms (whether to use acid concentration or base concentration) can be applied.