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
When gaseous nitrosyl chloride, NOCl (g), decomposes, the following equilibrium is established:
\[{\text{2NOCl(g)}} \rightleftharpoons {\text{2NO(g)}} + {\text{C}}{{\text{l}}_2}{\text{(g)}}\]
2.0 mol of NOCl(g) were placed in a \({\text{1.0 d}}{{\text{m}}^{\text{3}}}\) container and allowed to reach equilibrium. At equilibrium 1.0 mol of NOCl(g) was present. What is the value of \({K_{\text{c}}}\)?
A. 0.50
B. 1.0
C. 1.5
D. 2.0
▶️Answer/Explanation
A
\[{\text{2NOCl(g)}} \rightleftharpoons {\text{2NO(g)}} + {\text{C}}{{\text{l}}_2}{\text{(g)}}\]
Initially 2 moles of NOCl were there. Since at equilibrium 1 moles of NOCl are remaining it means 1 mol of NOCl has reacted, which would produce 1 mol NO and 0.5 moles Cl2 at equilibrium.
Kc = [NO]2[Cl2]/[NOCl]2
Volume is 1 dm3.
Kc = [1]2[0.5]/[1]2 = 0.5
Question
What is the relationship between \({\text{p}}{K_{\text{a}}}\), \({\text{p}}{K_{\text{b}}}\) and \({\text{p}}{K_{\text{w}}}\) for a conjugate acid–base pair?
A. \({\text{p}}{K_{\text{a}}} = {\text{p}}{K_{\text{w}}} + {\text{p}}{K_{\text{b}}}\)
B. \({\text{p}}{K_{\text{a}}} = {\text{p}}{K_{\text{w}}} – {\text{p}}{K_{\text{b}}}\)
C. \({\text{p}}{K_{\text{a}}} \times {\text{p}}{K_{\text{b}}} = {\text{p}}{K_{\text{w}}}\)
D. \(\frac{{{\text{p}}{K_{\text{a}}}}}{{{\text{p}}{K_{\text{b}}}}} = {\text{p}}{K_{\text{w}}}\)
▶️Answer/Explanation
B
\({\text{p}}{K_{\text{a}}} +{\text{p}}{K_{\text{b}}} = {\text{p}}{K_{\text{w}}}\)
So, \({\text{p}}{K_{\text{a}}} = {\text{p}}{K_{\text{w}}} – {\text{p}}{K_{\text{b}}}\)
Question
A mixture of 2.0 mol of \({{\text{H}}_{\text{2}}}\) and 2.0 mol of \({{\text{I}}_{\text{2}}}\) is allowed to reach equilibrium in the gaseous state at a certain temperature in a \({\text{1.0 d}}{{\text{m}}^{\text{3}}}\) flask. At equilibrium, 3.0 mol of HI are present. What is the value of \({K_{\text{c}}}\) for this reaction?
\[{{\text{H}}_{\text{2}}}{\text{(g)}} + {{\text{I}}_{\text{2}}}{\text{(g)}} \rightleftharpoons {\text{2HI(g)}}\]
A. \({K_{\text{c}}} = \frac{{{{(3.0)}^2}}}{{{{(0.5)}^2}}}\)
B. \({K_{\text{c}}} = \frac{{3.0}}{{{{(0.5)}^2}}}\)
C. \({K_{\text{c}}} = \frac{{{{(3.0)}^2}}}{{{{(2.0)}^2}}}\)
D. \({K_{\text{c}}} = \frac{{{{(0.5)}^2}}}{{{{(3.0)}^2}}}\)
▶️Answer/Explanation
A
At equilibrium 3 moles of HI are formed which means 1.5 moles of each H2 and I2 are consumed.
Initially, 2 moles each of H2 and I2 were there. So, at equilibrium, o.5 moles each of H2 and I2 are remaining. Volume = 1 dm3
Kc = [HI]2/[H2][I2]
\({K_{\text{c}}} = \frac{{{{(3.0)}^2}}}{{{{(0.5)}^2}}}\)
Question
The equation for the reaction between two gases, A and B, is:
\[{\text{2A(g)}} + {\text{3B(g)}} \rightleftharpoons {\text{C(g)}} + {\text{3D(g)}}\]
When the reaction is at equilibrium at 600 K the concentrations of A, B, C and D are 2, 1, 3 and 2 \({\text{mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) respectively. What is the value of the equilibrium constant at 600 K?
A. \(\frac{1}{6}\)
B. \(\frac{9}{7}\)
C. 3
D. 6
▶️Answer/Explanation
D
Kc = [C][D]3/[A]2[B]3
Kc = [3][2]3/[2]2[1]3
Kc = 6