Question
What is the approximate pH of a \({\text{0.01 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) ammonia solution?
A. 2
B. More than 2 but less than 7
C. More than 7 but less than 12
D. 12
▶️Answer/Explanation
C
NH3+H2O⇌NH4+ + OH−
Since Ammonia is a weak base, it does not dissociate completely in water.
If not completely dissociated, [OH−] formed would be less than 0.01 mol/dm3.
pOH = -Log[OH−],
pOH >2, pH <12. And since it is a base, pH>7.
Question
\({\text{100 c}}{{\text{m}}^{\text{3}}}\) of a NaOH solution of pH 12 is mixed with \({\text{900 c}}{{\text{m}}^{\text{3}}}\) of water. What is the pH of the resulting solution?
A. 1
B. 3
C. 11
D. 13
▶️Answer/Explanation
C
pOH = -Log[OH−]
pH+pOH =14
pOH = 14-12=2
[OH−] = 10-2 mol/l in 100cm3 or 0.1 l NaOH, i.e. 10-3moles NaOH are mixed with 900cm3 water.
Volume of final solution =1000cm3 or 1ltr having 10-3moles NaOH.
[NaOH]=10-11moles/ltr
pOH = -Log[OH−] = 3
pH = 14-3=11
Question
The indicator, HIn is used in a titration between an acid and base. Which statement about the dissociation of the indicator, HIn is correct?
\[{\text{HIn(aq)}} \rightleftharpoons {{\text{H}}^ + }{\text{(aq)}} + {\text{I}}{{\text{n}}^ – }{\text{(aq)}}\]
colour A colour B
A. In a strongly alkaline solution, colour B would be observed.
B. In a strongly acidic solution, colour B would be observed.
C. \({\text{[I}}{{\text{n}}^ – }{\text{]}}\) is greater than [HIn] at the equivalence point.
D. In a weakly acidic solution colour B would be observed.
▶️Answer/Explanation
A
Option A: In a strongly alkaline solution hydroxide ions will absorb H+ ions, hence equilibrium will shift towards forward direction and colour B of In– will be observed here.
Question
At the same concentration, which acid would have the lowest pH?
\(\begin{array}{*{20}{l}} {{\text{A.}}}&{{\text{HN}}{{\text{O}}_{\text{2}}}}&{{K_{\text{a}}} = 5.6 \times {{10}^{ – 4}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ – 3}}} \\ {{\text{B.}}}&{{\text{HF}}}&{{K_{\text{a}}} = 6.8 \times {{10}^{ – 4}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ – 3}}} \\ {{\text{C.}}}&{{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}}&{{K_{\text{a}}} = 6.3 \times {{10}^{ – 5}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ – 3}}} \\ {{\text{D.}}}&{{\text{HCN}}}&{{K_{\text{a}}} = 4.9 \times {{10}^{ – 10}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ – 3}}} \end{array}\)
▶️Answer/Explanation
B
The acid with highest Ka will be strongest and will have highest H+ concentration, hence lowest pH.
Question
Ammonia acts as a weak base when it reacts with water. What is the \({K_{\text{b}}}\) expression for this reaction?
A. \(\frac{{[{\text{NH}}_4^ + ][{\text{O}}{{\text{H}}^ – }]}}{{[{\text{N}}{{\text{H}}_3}][{{\text{H}}_2}{\text{O]}}}}\)
B. \(\frac{{[{\text{N}}{{\text{H}}_3}{\text{][}}{{\text{H}}_2}{\text{O]}}}}{{{\text{[NH}}_4^ + {\text{][O}}{{\text{H}}^ – }{\text{]}}}}\)
C. \(\frac{{[{\text{N}}{{\text{H}}_3}]}}{{[{\text{NH}}_4^ + ][{\text{O}}{{\text{H}}^ – }]}}\)
D. \(\frac{{[{\text{NH}}_4^ + ][{\text{O}}{{\text{H}}^ – }]}}{{[{\text{N}}{{\text{H}}_3}]}}\)
▶️Answer/Explanation
D
NH3+H2O⇌NH4++OH−
\({K_{\text{b}}}\) = \(\frac{{[{\text{NH}}_4^ + ][{\text{O}}{{\text{H}}^ – }]}}{{[{\text{N}}{{\text{H}}_3}]}}\)