Q. What is the effect of adding an inert gas at constant volume on the equilibrium of the reaction A(g) ⇌ B(g)?
A.
Shifts to the right
B.
Shifts to the left
C.
No effect
D.
Increases the rate of reaction
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Solution
Adding an inert gas at constant volume does not change the partial pressures of the reactants or products, hence no effect on equilibrium.
Correct Answer: C — No effect
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Q. What is the effect of dilution on the pH of a strong acid solution?
A.
pH increases
B.
pH decreases
C.
pH remains constant
D.
pH becomes neutral
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Solution
Diluting a strong acid decreases its concentration, which increases the pH (making it less acidic).
Correct Answer: A — pH increases
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Q. What is the effect of dilution on the pH of a strong acid?
A.
pH increases
B.
pH decreases
C.
pH remains constant
D.
pH becomes neutral
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Solution
Dilution of a strong acid decreases its concentration, thus increasing the pH.
Correct Answer: A — pH increases
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Q. What is the effect of dilution on the pH of a weak acid solution?
A.
pH decreases
B.
pH increases
C.
pH remains constant
D.
pH becomes neutral
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Solution
Dilution of a weak acid decreases its concentration, which shifts the equilibrium to the right, increasing the concentration of H+ ions and thus increasing the pH.
Correct Answer: B — pH increases
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Q. What is the effect of increasing temperature on an exothermic reaction at equilibrium?
A.
Shift to the right
B.
Shift to the left
C.
No change
D.
Increase the rate of reaction
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Solution
For an exothermic reaction, increasing the temperature shifts the equilibrium to the left, favoring the reactants.
Correct Answer: B — Shift to the left
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Q. What is the effect of increasing the distance from the pivot point on the torque produced by a force?
A.
Torque increases
B.
Torque decreases
C.
Torque remains the same
D.
Torque becomes zero
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Solution
Increasing the distance from the pivot point increases the torque produced by the force.
Correct Answer: A — Torque increases
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Q. What is the effect of increasing the pressure on the equilibrium of the reaction: 2N2(g) + 3H2(g) ⇌ 2NH3(g)?
A.
Shifts to the left
B.
Shifts to the right
C.
No effect
D.
Increases the rate of reaction
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Solution
Increasing the pressure will shift the equilibrium to the right, favoring the formation of NH3, as there are fewer moles of gas on the product side.
Correct Answer: B — Shifts to the right
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Q. What is the effect of temperature on the equilibrium constant for an exothermic reaction?
A.
Increases with temperature
B.
Decreases with temperature
C.
Remains constant
D.
Depends on the concentration
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Solution
For exothermic reactions, increasing the temperature shifts the equilibrium to the left, thus decreasing the equilibrium constant.
Correct Answer: B — Decreases with temperature
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Q. What is the effect of temperature on the equilibrium constant K for an exothermic reaction?
A.
K increases with temperature
B.
K decreases with temperature
C.
K remains constant
D.
K is independent of temperature
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Solution
For an exothermic reaction, increasing the temperature shifts the equilibrium to the left, thus decreasing the value of K.
Correct Answer: B — K decreases with temperature
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Q. What is the equilibrium constant expression for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g)?
A.
Kc = [NH3]^2 / ([N2][H2]^3)
B.
Kc = [N2][H2]^3 / [NH3]^2
C.
Kc = [NH3]^2 / [N2][H2]
D.
Kc = [N2][H2] / [NH3]^2
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Solution
The equilibrium constant Kc is given by the ratio of the concentration of products to reactants, raised to the power of their coefficients.
Correct Answer: A — Kc = [NH3]^2 / ([N2][H2]^3)
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Q. What is the equilibrium constant expression for the reaction: 2A + B ⇌ C?
A.
[C]/([A]^2[B])
B.
[A]^2[B]/[C]
C.
[C]/[A][B]
D.
[A][B]/[C]
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Solution
The equilibrium constant K is given by the expression K = [C]/([A]^2[B]) for the reaction 2A + B ⇌ C.
Correct Answer: A — [C]/([A]^2[B])
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Q. What is the equilibrium constant expression for the reaction: aA + bB ⇌ cC + dD?
A.
K = [C]^c [D]^d / [A]^a [B]^b
B.
K = [A]^a [B]^b / [C]^c [D]^d
C.
K = [C]^c [D]^d
D.
K = [A]^a [B]^b
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Solution
The equilibrium constant K is defined as the ratio of the concentration of products to the concentration of reactants, each raised to the power of their coefficients in the balanced equation.
Correct Answer: A — K = [C]^c [D]^d / [A]^a [B]^b
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Q. What is the Kb of ammonia (NH3) if the pKa of its conjugate acid (NH4+) is 9.25?
A.
1.8 x 10^-5
B.
5.6 x 10^-10
C.
1.0 x 10^-14
D.
3.2 x 10^-5
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Solution
Kb = Kw / Ka = 1.0 x 10^-14 / 5.6 x 10^-10 = 1.8 x 10^-5
Correct Answer: A — 1.8 x 10^-5
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Q. What is the Ksp expression for the salt Ag2SO4?
A.
Ksp = [Ag+]^2[SO4^2-]
B.
Ksp = [Ag2+]^2[SO4^2-]
C.
Ksp = [Ag+]^2[SO4^2-]^2
D.
Ksp = [Ag+]^2[SO4^2-]^3
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Solution
The solubility product constant (Ksp) for Ag2SO4 is given by Ksp = [Ag+]^2[SO4^2-] because the dissociation is Ag2SO4 ⇌ 2Ag+ + SO4^2-.
Correct Answer: A — Ksp = [Ag+]^2[SO4^2-]
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Q. What is the Ksp of AgCl if the solubility of AgCl in water is 1.0 x 10^-5 M?
A.
1.0 x 10^-10
B.
1.0 x 10^-5
C.
1.0 x 10^-15
D.
1.0 x 10^-20
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Solution
Ksp = [Ag+][Cl-] = (1.0 x 10^-5)(1.0 x 10^-5) = 1.0 x 10^-10.
Correct Answer: A — 1.0 x 10^-10
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Q. What is the pH of a 0.01 M NaOH solution?
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Solution
For NaOH, which is a strong base, pOH = -log[OH-]. [OH-] = 0.01 M, so pOH = 2. Therefore, pH = 14 - pOH = 14 - 2 = 12.
Correct Answer: A — 12
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Q. What is the pH of a 0.01 M solution of sodium acetate (CH3COONa)?
A.
4.76
B.
7.00
C.
9.24
D.
10.00
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Solution
Sodium acetate is a salt of a weak acid (acetic acid) and a strong base (sodium hydroxide). The pH can be calculated using the formula pH = 7 + 0.5(pKa - log[C]), where pKa of acetic acid is 4.76.
Correct Answer: C — 9.24
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Q. What is the pH of a 0.01 M solution of sodium hydroxide (NaOH)?
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Solution
pH = 14 - pOH; pOH = -log[OH-] = -log(0.01) = 2; pH = 14 - 2 = 12.
Correct Answer: A — 12
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Q. What is the pH of a 0.01 M solution of sulfuric acid (H2SO4)?
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Solution
H2SO4 is a strong acid and dissociates completely, so pH = -log(0.01) = 2.
Correct Answer: A — 1
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Q. What is the pH of a 0.05 M solution of sulfuric acid (H2SO4)?
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Solution
H2SO4 is a strong acid and dissociates completely; [H+] = 0.05 M, so pH = -log(0.05) ≈ 1.3
Correct Answer: B — 1.3
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Q. What is the pH of a 0.1 M NaOH solution?
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Solution
pOH = -log[OH-] = -log(0.1) = 1; pH = 14 - pOH = 14 - 1 = 13
Correct Answer: C — 12
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Q. What is the pH of a 0.1 M solution of acetic acid (CH3COOH) given its Ka is 1.8 x 10^-5?
A.
2.87
B.
4.76
C.
3.87
D.
5.00
Show solution
Solution
Using the formula for weak acids, pH = 0.5(pKa - log[C]) = 0.5(4.74 - log(0.1)) = 4.76.
Correct Answer: B — 4.76
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Q. What is the pH of a 0.1 M solution of ammonium chloride (NH4Cl)?
A.
5.1
B.
5.5
C.
6.1
D.
6.5
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Solution
NH4Cl is a salt of a weak base and a strong acid; it hydrolyzes to give H+, resulting in a pH < 7.
Correct Answer: C — 6.1
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Q. What is the pH of a 0.1 M solution of hydrochloric acid (HCl)?
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Solution
The pH of a strong acid like HCl is calculated using the formula pH = -log[H+]. For 0.1 M HCl, [H+] = 0.1 M, so pH = -log(0.1) = 1.
Correct Answer: A — 1
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Q. What is the pH of a 0.1 M solution of potassium hydrogen phthalate (KHP)?
A.
4.0
B.
5.0
C.
6.0
D.
7.0
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Solution
KHP is a weak acid; its pKa is approximately 5.0, so the pH of a 0.1 M solution is around 5.0.
Correct Answer: B — 5.0
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Q. What is the pH of a 0.1 M solution of sodium acetate (CH3COONa)?
A.
4.76
B.
7
C.
9.24
D.
10
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Solution
pH = 7 + 0.5(pKa + log[C]) = 7 + 0.5(4.76 + log(0.1)) = 9.24
Correct Answer: C — 9.24
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Q. What is the pH of a 0.1 M solution of sodium bicarbonate (NaHCO3)?
A.
7.5
B.
8.4
C.
9.0
D.
6.0
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Solution
NaHCO3 is a weak base; its pH is around 8.4.
Correct Answer: B — 8.4
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Q. What is the pH of a buffer solution containing 0.1 M acetic acid and 0.1 M sodium acetate?
A.
4.74
B.
5.74
C.
6.74
D.
7.74
Show solution
Solution
Using the Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]). Here, pKa ≈ 4.74, so pH = 4.74 + log(1) = 4.74.
Correct Answer: A — 4.74
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Q. What is the pH of a buffer solution containing 0.2 M acetic acid and 0.1 M sodium acetate?
A.
4.76
B.
5.00
C.
5.74
D.
6.00
Show solution
Solution
Using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) = 4.76 + log(0.1/0.2) = 5.74
Correct Answer: C — 5.74
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Q. What is the pH of a buffer solution made from 0.2 M acetic acid and 0.2 M sodium acetate?
A.
4.76
B.
5.76
C.
6.76
D.
7.76
Show solution
Solution
Using the Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]) = 4.76.
Correct Answer: A — 4.76
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