Question

Dissociation constants of \[C{{H}_{3}}COOH\] and \[N{{H}_{4}}OH\] in an aqueous solution are 10-5. If the pH of \[C{{H}_{3}}COOH\] solution is 3 what will be the pH of \[N{{H}_{4}}OH\]?
(A) 3.0
(B) 4.0
(C) 10.0
(D) 11.0

Answer 1

Hint: Remember that the relationship between the pH and the pOH of a solution is given as follows:
\[pOH\text{ }=\text{ }14\text{ }\text{ }pH\]
Also, the relationship between the dissociation constant ${{K}_{a}}$, the molarity of the given solution and the presence of hydronium ions if given by the following equation:
\[[{{H}_{3}}{{O}^{+}}]=\sqrt{c\times {{K}_{a}}}\]

Step-by-Step Solution:
Let us first understand what the pH of a solution and its dissociation constants really mean, before moving on to the specifics of this question.
The pH of a solution is a measure of the molar concentration of hydrogen ions in the solution and as such is a measure of the acidity or basicity of the solution. The letters pH stands for the "power of hydrogen" and the numerical value is defined as the negative base 10 logarithm of the molar concentration of hydrogen ions.
\[pH\text{ }=\text{ }-lo{{g}_{10}}\left[ {{H}^{+}} \right]\]
The usual range of pH values encountered is between 0 and 14, with 0 being the value for concentrated hydrochloric acid (1 M HCl), 7 the value for pure water (neutral pH), and 14 being the value for concentrated sodium hydroxide (1 M NaOH).
The acid dissociation constant is the equilibrium constant of the dissociation reaction of an acid and is denoted by ${{K}_{a}}$. This equilibrium constant is a quantitative measure of the strength of an acid in a solution. ${{K}_{a}}$ is commonly expressed in units of mol/L.
Now, let us try and solve the given question.
\[\]pH of \[C{{H}_{3}}COOH\] sol = 3
\[\therefore [{{H}_{3}}{{O}^{+}}]=\sqrt{c\times {{K}_{a}}}={{10}^{-3}}\]
\[\Rightarrow {{10}^{-3}}=\sqrt{c\times {{10}^{-5}}}\]
\[\Rightarrow {{10}^{-6}}=c\times {{10}^{-5}}\]
 \[\Rightarrow {{10}^{-1}}mol/l=c \]
 Given,
 \[{{K}_{a}}={{K}_{b}}={{10}^{-5}} \]
Therefore, the pH of a $10^{-3}$ solution of \[N{{H}_{4}}OH\] can be calculated by:
 \[\therefore [O{{H}^{-}}]=\sqrt{c\times {{K}_{b}}} \]
 \[\Rightarrow [O{{H}^{-}}]=\sqrt{{{10}^{-1}}\times {{10}^{-5}}} \]
 \[\Rightarrow [O{{H}^{-}}]={{10}^{-3}} \]
 \[\because pOH\text{ }=\text{ }-lo{{g}_{10}}\left[ O{{H}^{-}} \right] \]
 \[\Rightarrow pOH\text{ }=\text{ }3 \]
\[\therefore pH=14-pOH=11 \]

Thus, we can conclude that the answer to this question is d) 11.0

Note: There are two main methods of solving for hydrogen ion concentration. One involves the quadratic equation. The other assumes the weak acid barely dissociates in water and approximates the pH. Which one you choose depends on how accurate you need the answer to be.