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Q. No.Which of the following happens when NH4OH is added gradually to the solution containing 1M A2+ and 1M B3+ ions?
Given: \(\mathrm{K}_{\mathrm{sp}}\left[\mathrm{~A}(\mathrm{OH})_2\right]=9 \times 10^{-10} \text {}\)
\(\mathrm{K}_{\mathrm{sp}}\left[\mathrm{~B}(\mathrm{OH})_3\right]=27 \times 10^{-18} \text { at } 298 \mathrm{~K} .\)
1.
B(OH)3 will precipitate before A(OH)2
2.
A(OH)2 and B(OH)3 will precipitate together
3.
A(OH)2 will precipitate before B(OH)3
4.
Both A(OH)2 and B(OH)3 do not show precipitation with NH4OH
Subtopic: Solubility Product | Common Ion Effect |
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If 1 mM solution of ethylamine produces pH = 9, then the ionization constant (Kb) of ethylamine is 10–x.
The value of x is: (nearest integer)
[Note: The degree of ionization of ethylamine can be neglected with respect to unity]
1. 5
2. 7
3. 9
4. 11
The value of x is: (nearest integer)
[Note: The degree of ionization of ethylamine can be neglected with respect to unity]
1. 5
2. 7
3. 9
4. 11
Subtopic: Ionisation Constant of Acid, Base & Salt | pH calculation |
69%
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For a weak acid HA with a degree of dissociation x, which of the following gives the correct expression for \((\text{pH} - \text{pKa})\)?
| 1. | log (1+2x) | 2. | \(\log \left(\dfrac{1-x}{x}\right)\) |
| 3. | 0 | 4. | \(\log \left(\dfrac{x}{1-x}\right)\) |
Subtopic: Ionisation Constant of Acid, Base & Salt |
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Consider the given reaction:
\(\mathrm{H}_2(\mathrm{g})+\mathrm{I}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{~g})\)
If initially only \(\mathrm{H_2}\) and \(\mathrm{I_2}\) are present, which graph correctly predicts the attainment of equilibrium?
\(\mathrm{H}_2(\mathrm{g})+\mathrm{I}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{~g})\)
If initially only \(\mathrm{H_2}\) and \(\mathrm{I_2}\) are present, which graph correctly predicts the attainment of equilibrium?
| 1. |  |
2. |  |
| 3. |  |
4. |  |
Subtopic: Introduction To Equilibrium |
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Level 1: 80%+
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Determine the final concentration of the weak acid \(\text {HX}\) (given \(\mathrm{K}_{\mathrm{a}}=4 \times 10^{-10}\)) after a \(0.01~ \mathrm{M}\) solution has been diluted to achieve a \(\mathrm{pH}=6\). Express this final concentration in the form \(\mathrm{x \times 10^{-4} M}\), and then state the numerical value of x:
1. 14
2. 25
3. 31
4. 12
1. 14
2. 25
3. 31
4. 12
Subtopic: pH calculation |
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Which of the following is true for the degree of dissociation \(\alpha\) and pressure \(\mathrm{{P}_{T}}, \) for the reversible reaction \(\mathrm{{A}({g})} \rightleftharpoons \mathrm{{B}({g})+{C}({g})} \)?
1. If \(\mathrm{{P}_{T} \gg K_{P}},\) then \(\alpha \approx 1 \)
2. If \(\mathrm{{P}_{T}}\) increases, then \(\alpha\) decreases
3. If \(\mathrm{{P}_{T}}\) increases, then \(\alpha\) increases
4. If \(\mathrm{K_P \gg P_T,} \) then \(\alpha \) tends towards 0
1. If \(\mathrm{{P}_{T} \gg K_{P}},\) then \(\alpha \approx 1 \)
2. If \(\mathrm{{P}_{T}}\) increases, then \(\alpha\) decreases
3. If \(\mathrm{{P}_{T}}\) increases, then \(\alpha\) increases
4. If \(\mathrm{K_P \gg P_T,} \) then \(\alpha \) tends towards 0
Subtopic: Le Chatelier's principle |
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The pH of pure water is 7.0 at 25°C. When the temperature of pure water is increased to 80°C, what happens to its pH value?
1. The pH decreases
2. The pH remains the same (still 7.0)
3. The pH first increases and then decreases
4. The pH increases
Subtopic: Introduction To Equilibrium | pH calculation |
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Calculate the equilibrium constant Kp for the reaction
2N₂O₅(g) ⇌ 2N₂O₄(g) + O₂(g)
at 500 K, given that 37.8 g of N2O5 was placed in a 1 L reaction vessel, and the total pressure at equilibrium was found to be 18.65 bar.
[Assume N2O5 behaves ideally under these conditions]
[Given: R = 0.082 bar L mol–1 K–1 ]
1. 96.2
2. 6.75
3. 9.619
4. 18.6
2N₂O₅(g) ⇌ 2N₂O₄(g) + O₂(g)
at 500 K, given that 37.8 g of N2O5 was placed in a 1 L reaction vessel, and the total pressure at equilibrium was found to be 18.65 bar.
[Assume N2O5 behaves ideally under these conditions]
[Given: R = 0.082 bar L mol–1 K–1 ]
1. 96.2
2. 6.75
3. 9.619
4. 18.6
Subtopic: Kp, Kc & Factors Affecting them |
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Consider the reaction:
\(\mathrm{X}_2 \mathrm{Y}(\mathrm{~g}) \rightleftharpoons \mathrm{X}_2(\mathrm{~g})+\frac{1}{2} \mathrm{Y}_2(\mathrm{~g})\)
The equation representing the correct relationship between the degree of dissociation (x) of X2Y(g) with its equilibrium constant Kp is:
(Assume x to be very very small)
\(\mathrm{X}_2 \mathrm{Y}(\mathrm{~g}) \rightleftharpoons \mathrm{X}_2(\mathrm{~g})+\frac{1}{2} \mathrm{Y}_2(\mathrm{~g})\)
The equation representing the correct relationship between the degree of dissociation (x) of X2Y(g) with its equilibrium constant Kp is:
(Assume x to be very very small)
| 1. | \(\mathrm{x}=\sqrt[3]{\dfrac{2 \mathrm{Kp}}{\mathrm{p}}}\) | 2. | \(x=\sqrt[3]{\dfrac{2 \mathrm{Kp}^2}{\mathrm{p}}}\) |
| 3. | \(x=\sqrt[3]{\dfrac{\mathrm{Kp}}{2 \mathrm{p}}}\) | 4. | \(x=\sqrt[3]{\dfrac{K p}{p}}\) |
Subtopic: Kp, Kc & Factors Affecting them | Ionisation Constant of Acid, Base & Salt |
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\(\text K_{sp}\) for Cr(OH)3 is 1.6 × 10–30. What is the molar solubility of this salt in water?
| 1. | \(\sqrt[4]{\dfrac{1.6 \times 10^{-30}}{27}}\) | 2. | \(\dfrac{1.8 \times 10^{-30}}{27}\) |
| 3. | \(\sqrt[5]{1.8 \times 10^{-30}} \) | 4. | \(\sqrt[2]{1.6 \times 10^{-30}}\) |
Subtopic: Solubility Product |
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Level 1: 80%+
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