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Question: Given,\({K_a}\) for \(C{H_3}COOH\) is \(1.8 \times {10^{ - 5}}\) and \({K_b}\) for \(N{H_4}OH\) is \...

Given,Ka{K_a} for CH3COOHC{H_3}COOH is 1.8×1051.8 \times {10^{ - 5}} and Kb{K_b} for NH4OHN{H_4}OH is 1.8×1051.8 \times {10^{ - 5}}.The pHpH of ammonium acetate will be:
A. 7.0057.005
B. 4.754.75
C. 7.07.0
D. between 66 and 77

Explanation

Solution

Ammonium acetate is the salt formed between the reaction of acetic acid (CH3COOHC{H_3}COOH), an acid, and ammonium hydroxide (NH4OHN{H_4}OH), a base. Thus, the pHpH of ammonium acetate will be related to the dissociation constants of the acid and base, and is given by the pHpH equation.
Formulas used: pH=7+12[pKapKb]pH = 7 + \dfrac{1}{2}\left[ {p{K_a} - p{K_b}} \right]
Where pKap{K_a} denotes the negative log of the acid dissociation constant (Ka{K_a}) and pKbp{K_{\mathbf{b}}} denotes the negative log of the base dissociation constant (Kb{K_b}).
That is, pKa=log(Ka)p{K_a} = - \log ({K_a}) and pKa=logKbp{K_a} = - \log {K_b}

Complete step by step answer:
Ammonium acetate is the salt formed due to the neutralization reaction between acetic acid and ammonium hydroxide. The reaction is as follows:
CH3COOH+NH4OHCH3COONH4+H2OC{H_3}COOH + N{H_4}OH \rightleftharpoons C{H_3}COON{H_4} + {H_2}O
Thus, the pHpH of this solution can be given by the following equation:
pH=7+12[pKapKb]pH = 7 + \dfrac{1}{2}\left[ {p{K_a} - p{K_b}} \right]
Where pKap{K_a} denotes the negative log of the acid dissociation constant (Ka{K_a}) and pKbp{K_{\mathbf{b}}} denotes the negative log of the base dissociation constant (Kb{K_b}).
Here, in this question, we have the acid and base dissociation constants to be of equal value, since:
Ka=Kb=1.8×105{K_a} = {K_b} = 1.8 \times {10^{ - 5}}
Thus, if we take the negative logarithms on both sides,
log(Ka)=log(Kb)- \log ({K_a}) = - \log ({K_b})
But as we know, the negative logarithm of the acid dissociation constant is equal to pKap{K_a} and the negative logarithm of the base dissociation constant is equal to pKbp{K_b}. Hence, in this question:
pKa=pKbpKapKb=0p{K_a} = p{K_b} \Rightarrow p{K_a} - p{K_b} = 0
Substituting this value in our equation for pHpH, we get:
pH=7+12[0]=7pH = 7 + \dfrac{1}{2}\left[ 0 \right] = 7
Hence, the pHpH of the resulting ammonium acetate solution will be 7.07.0.
Hence, the correct option to be marked is option C.

Note: The acid and base dissociation constants represent the relative strength of the acid and base respectively. Larger the value of Ka/Kb{K_a}/{K_b}, stronger will be the acid/base.
pHpH stands for potential of hydrogen, and measures the concentration of hydrogen ions (H+{H^ + }) in a solution. pHpH of less than 77 indicates an acidic solution, while more than 77 indicates a basic solution. Note that the 77, written as the first term in the pHpH equation, indicates the pHpH of pure water.