Acids, Bases, and Electrochemistry
Acids, Bases, and Electrochemistry MCQ Questions
7.
The pH of a solution is 5. What is the hydroxide ion concentration?
Approach
To find the hydroxide ion concentration (\([\text{OH}^-]\)) from the pH of a solution, we first need to calculate the hydrogen ion concentration (\([\text{H}^+]\)) using the pH definition. Then, we use the ion product of water (\(K_w\)) to determine the hydroxide ion concentration.
Step-by-step
- Calculate \([\text{H}^+]\) from pH:
The pH of a solution is defined as the negative logarithm (base 10) of the hydrogen ion concentration:
\[ \text{pH} = -\log[\text{H}^+] \] Given \(\text{pH} = 5\), we can write:
\[ 5 = -\log[\text{H}^+] \] Multiplying by -1:
\[ -5 = \log[\text{H}^+] \] To find \([\text{H}^+]\), we take the antilog (base 10) of -5:
\[ [\text{H}^+] = 10^{-5} \text{ M} \] - Calculate \([\text{OH}^-]\) using the ion product of water (\(K_w\)):
At \(25^\circ\text{C}\) (standard temperature), the ion product of water is a constant:
\[ K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} \text{ M}^2 \] We have \([\text{H}^+] = 10^{-5} \text{ M}\). Substitute this value into the \(K_w\) expression:
\[ (10^{-5} \text{ M})[\text{OH}^-] = 10^{-14} \text{ M}^2 \] Now, solve for \([\text{OH}^-]\):
\[ [\text{OH}^-] = \frac{10^{-14} \text{ M}^2}{10^{-5} \text{ M}} \]\[ [\text{OH}^-] = 10^{(-14) - (-5)} \text{ M} \]\[ [\text{OH}^-] = 10^{-14 + 5} \text{ M} \]\[ [\text{OH}^-] = 10^{-9} \text{ M} \]
Correct Option
C) \(10^{-9} \text{ M}\) is the calculated hydroxide ion concentration.
Incorrect Options
- A) \(10^{-7} \text{ M}\) — This would be the concentration of both \([\text{H}^+]\) and \([\text{OH}^-]\) in a neutral solution (\(\text{pH} = 7\)).
- B) \(10^{-5} \text{ M}\) — This is the hydrogen ion concentration (\([\text{H}^+]\)), not the hydroxide ion concentration.
- D) \(10^{-14} \text{ M}\) — This value represents the ion product of water (\(K_w\)), not the hydroxide ion concentration.
8.
The pH of 10⁻⁸ M HCl solution is approximately:
Approach
When dealing with very dilute strong acid solutions (concentration less than or equal to \(10^{-7} \text{ M}\)), the contribution of \(\text{H}^+\) ions from the autoionization of water cannot be ignored. In such cases, the total \(\text{H}^+\) concentration is the sum of \(\text{H}^+\) from the acid and \(\text{H}^+\) from water.
Step-by-step
Identify the given concentration: The concentration of \(\text{HCl}\) is \(10^{-8} \text{ M}\). Since \(\text{HCl}\) is a strong acid, it dissociates completely:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]Therefore, the \(\text{H}^+\) concentration from \(\text{HCl}\) is \([\text{H}^+]_{\text{acid}} = 10^{-8} \text{ M}\).Consider autoionization of water: Water autoionizes to produce \(\text{H}^+\) and \(\text{OH}^-\) ions:
\[ \text{H}_2\text{O}
ightleftharpoons \text{H}^+ + \text{OH}^- \]At \(25^\circ\text{C}\), the ion product of water, \(\text{K}_{\text{w}}\), is \(10^{-14}\):\[ \text{K}_{\text{w}} = [\text{H}^+][\text{OH}^-] = 10^{-14} \]Set up equilibrium for water: Let \([\text{H}^+]_{\text{water}}\) be the concentration of \(\text{H}^+\) ions produced from water. Due to the common ion effect, the autoionization of water will be suppressed by the \(\text{H}^+\) from \(\text{HCl}\). Let the equilibrium concentration of \(\text{OH}^-\) be \([\text{OH}^-]\). Then, the total \(\text{H}^+\) concentration at equilibrium will be \([\text{H}^+]_{\text{total}} = [\text{H}^+]_{\text{acid}} + [\text{H}^+]_{\text{water}} = 10^{-8} + [\text{OH}^-]\). (Since for every \(\text{OH}^-\) produced by water, an \(\text{H}^+\) is also produced, \([\text{H}^+]_{\text{water}} = [\text{OH}^-]\) from water's dissociation).
So, \([\text{H}^+]_{\text{total}} = 10^{-8} + [\text{OH}^-]\).
Substitute into \(\text{K}_{\text{w}}\) expression:
\[ (10^{-8} + [\text{OH}^-])[\text{OH}^-] = 10^{-14} \]Let \(x = [\text{OH}^-]\):\[ (10^{-8} + x)x = 10^{-14} \]\[ x^2 + 10^{-8}x - 10^{-14} = 0 \]Solve the quadratic equation for \(x\): Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ x = \frac{-10^{-8} \pm \sqrt{(10^{-8})^2 - 4(1)(-10^{-14})}}{2(1)} \]\[ x = \frac{-10^{-8} \pm \sqrt{10^{-16} + 4 \times 10^{-14}}}{2} \]\[ x = \frac{-10^{-8} \pm \sqrt{10^{-16} + 400 \times 10^{-16}}}{2} \]\[ x = \frac{-10^{-8} \pm \sqrt{401 \times 10^{-16}}}{2} \]\[ x = \frac{-10^{-8} \pm \sqrt{4.01 \times 10^{-14}}}{2} \]\[ x = \frac{-10^{-8} \pm 2.0025 \times 10^{-7}}{2} \]Since concentration cannot be negative, we take the positive root:\[ x = \frac{-10^{-8} + 2.0025 \times 10^{-7}}{2} \]\[ x = \frac{-0.1 \times 10^{-7} + 2.0025 \times 10^{-7}}{2} \]\[ x = \frac{1.9025 \times 10^{-7}}{2} \]\[ x = 9.5125 \times 10^{-8} \text{ M} \]So, \([\text{OH}^-] = 9.5125 \times 10^{-8} \text{ M}\).Calculate total \(\text{H}^+\) concentration:
\[ [\text{H}^+]_{\text{total}} = 10^{-8} + [\text{OH}^-] \]\[ [\text{H}^+]_{\text{total}} = 10^{-8} + 9.5125 \times 10^{-8} \]\[ [\text{H}^+]_{\text{total}} = (1 + 9.5125) \times 10^{-8} \]\[ [\text{H}^+]_{\text{total}} = 10.5125 \times 10^{-8} \]\[ [\text{H}^+]_{\text{total}} = 1.05125 \times 10^{-7} \text{ M} \]Calculate pH:
\[ \text{pH} = -\log_{10}[\text{H}^+]_{\text{total}} \]\[ \text{pH} = -\log_{10}(1.05125 \times 10^{-7}) \]\[ \text{pH} = -(\log_{10}(1.05125) + \log_{10}(10^{-7})) \]\[ \text{pH} = -(\log_{10}(1.05125) - 7) \]\[ \text{pH} = 7 - \log_{10}(1.05125) \]\[ \text{pH} \approx 7 - 0.021 \]\[ \text{pH} \approx 6.979 \]Rounding to two decimal places, \(\text{pH} \approx 6.98\).
Correct Option: C) 6.98
Incorrect Options
- A) 8 — This would be the pH if we simply calculated \(-\log(10^{-8})\) without considering the autoionization of water. However, an acid solution cannot have a pH greater than 7.
- B) 6 — This value is too low. If the pH were 6, the \([\text{H}^+]\) would be \(10^{-6} \text{ M}\), which is much higher than the \(\text{HCl}\) concentration.
- D) 7 — A pH of 7 indicates a neutral solution. While the \(\text{HCl}\) concentration is very low, it is still an acid, and thus its pH must be slightly less than 7, not exactly 7.
9.
Equal volumes of pH 3 and pH 5 strong acid solutions are mixed. The pH of the resulting solution is approximately:
Approach
To find the pH of the resulting solution when two strong acid solutions are mixed, we need to calculate the total number of moles of H⁺ ions and the total volume. Since pH is a logarithmic scale, we cannot simply average the pH values. We must convert pH to H⁺ concentration, calculate the total moles, and then convert back to pH.
Step-by-step
-
Calculate \([H^+]\) for pH 3 solution:
Given pH = 3, the concentration of H⁺ ions is:
\[ [H^+]_1 = 10^{-\text{pH}_1} = 10^{-3} \text{ M} \]
-
Calculate \([H^+]\) for pH 5 solution:
Given pH = 5, the concentration of H⁺ ions is:
\[ [H^+]_2 = 10^{-\text{pH}_2} = 10^{-5} \text{ M} \]
-
Assume equal volumes:
Let the volume of each solution be \(V\) liters. So, \(V_1 = V\) and \(V_2 = V\).
-
Calculate moles of H⁺ in each solution:
Moles of H⁺ in the first solution (\(n_1\)):
\[ n_1 = [H^+]_1 \times V_1 = 10^{-3} \times V \text{ moles} \]
Moles of H⁺ in the second solution (\(n_2\)):
\[ n_2 = [H^+]_2 \times V_2 = 10^{-5} \times V \text{ moles} \]
-
Calculate total moles of H⁺:
\[ n_{\text{total}} = n_1 + n_2 = (10^{-3} \times V) + (10^{-5} \times V) \]
\[ n_{\text{total}} = V (10^{-3} + 10^{-5}) = V (0.001 + 0.00001) = V (0.00101) \text{ moles} \]
-
Calculate total volume:
\[ V_{\text{total}} = V_1 + V_2 = V + V = 2V \text{ liters} \]
-
Calculate the new concentration of H⁺ in the mixed solution:
\[ [H^+]_{\text{new}} = \frac{n_{\text{total}}}{V_{\text{total}}} = \frac{V (0.00101)}{2V} = \frac{0.00101}{2} = 0.000505 \text{ M} \]
-
Calculate the pH of the resulting solution:
\[ \text{pH}_{\text{new}} = -\log_{10}([H^+]_{\text{new}}) = -\log_{10}(0.000505) \]
\[ \text{pH}_{\text{new}} \approx -\log_{10}(5.05 \times 10^{-4}) \]
\[ \text{pH}_{\text{new}} \approx -(\log_{10}(5.05) + \log_{10}(10^{-4})) \]
\[ \text{pH}_{\text{new}} \approx -(0.703 - 4) = -(-3.297) = 3.297 \]
Rounding to one decimal place, the pH is approximately 3.3.
Correct Option: A) 3.3
Incorrect Options
- B) 3.0 — This would be the pH if only the stronger acid (pH 3) were present, or if the contribution of the pH 5 solution was negligible, which is not entirely the case when equal volumes are mixed. The pH should increase slightly from 3.0 because of the dilution effect and the addition of a less concentrated acid.
- C) 4.0 — This would be the simple average of the pH values, which is incorrect because pH is a logarithmic scale and cannot be averaged directly.
- D) 5.0 — This would be the pH if only the weaker acid (pH 5) were present, or if the pH 3 solution was diluted so much that its effect became negligible, which is not the scenario here. The resulting pH must be between 3 and 5, and closer to the stronger acid's pH.
10.
Which of the following solutions is the most acidic?
ANSWER :
C. Solution with pH = 2
Approach
The acidity or alkalinity of a solution is measured by its pH value. The pH scale ranges from 0 to 14. A lower pH value indicates higher acidity, while a higher pH value indicates higher alkalinity (basicity). A pH of 7 is considered neutral.
Step-by-step
- Understand the pH scale: The pH scale is a logarithmic scale used to specify the acidity or basicity of an aqueous solution. It is defined as the negative base-10 logarithm of the hydrogen ion concentration \([\text{H}^+]\).
\[ \text{pH} = -\log_{10}[\text{H}^+] \] - Interpret pH values:
- Solutions with \(\text{pH} < 7\) are acidic.
- Solutions with \(\text{pH} = 7\) are neutral.
- Solutions with \(\text{pH} > 7\) are basic (alkaline).
- Relate pH to acidity: The further a pH value is below 7, the more acidic the solution. This means that a solution with a pH of 1 is more acidic than a solution with a pH of 2, and so on.
- Compare the given options: We are given four pH values: 7, 10, 2, and 5. We need to find the most acidic solution.
- \(\text{pH} = 7\): Neutral
- \(\text{pH} = 10\): Basic
- \(\text{pH} = 2\): Acidic
- \(\text{pH} = 5\): Acidic
- Identify the most acidic: Among the acidic solutions (\(\text{pH}=2\) and \(\text{pH}=5\)), the solution with the lowest pH value is the most acidic. Therefore, a solution with \(\text{pH}=2\) is the most acidic among the given options.
Correct Option: C) Solution with pH = 2 is the most acidic because it has the lowest pH value among the given options, indicating the highest concentration of hydrogen ions.
Incorrect Options
- A) Solution with pH = 7 — This solution is neutral, meaning it is neither acidic nor basic.
- B) Solution with pH = 10 — This solution is basic (alkaline) because its pH value is greater than 7.
- D) Solution with pH = 5 — This solution is acidic, but it is less acidic than a solution with pH = 2. A lower pH indicates higher acidity.
11.
The value of Kw at 25°C is:
Approach
The question asks for the value of the ion product of water, \(K_w\), at a specific temperature, 25°C. This is a fundamental constant in chemistry, particularly in acid-base equilibria, and its value is standard at this temperature.
Step-by-step
- Understanding \(K_w\): The ion product of water, \(K_w\), represents the equilibrium constant for the autoionization of water. Water molecules dissociate into hydrogen ions (\(H^+\), often represented as hydronium ions, \(H_3O^+\)) and hydroxide ions (\(OH^-\)). The equilibrium reaction is:\[ H_2O_{(l)}
ightleftharpoons H^+_{(aq)} + OH^-_{(aq)} \]
- Expression for \(K_w\): The equilibrium constant for this reaction is given by the product of the molar concentrations of the ions:\[ K_w = [H^+][OH^-] \]
- Value at 25°C: At 25°C (or 298 K), which is considered standard room temperature in chemistry, the concentration of both \(H^+\) and \(OH^-\) in pure water is \(1 \times 10^{-7} \text{ M}\). This is because pure water is neutral, meaning \([H^+] = [OH^-]\).
- Calculation of \(K_w\): Substituting these values into the \(K_w\) expression:\[ K_w = (1 \times 10^{-7} \text{ M}) \times (1 \times 10^{-7} \text{ M}) \]\[ K_w = 1 \times 10^{-14} \text{ M}^2 \]The units are often omitted when stating \(K_w\) values.
Correct Option: D) The value of \(K_w\) at 25°C is a well-established constant, \(1 \times 10^{-14}\). This value is crucial for calculating pH and pOH of aqueous solutions at this temperature.
Incorrect Options
- A) 1 × 10⁻¹² — This value is incorrect for \(K_w\) at 25°C. It would imply a different equilibrium between \(H^+\) and \(OH^-\) than observed in pure water at this temperature.
- B) 1 × 10⁻⁷ — This is the concentration of either \([H^+]\) or \([OH^-]\) in pure water at 25°C, not the value of \(K_w\).
- C) 1 × 10⁻¹⁰ — This value is incorrect for \(K_w\) at 25°C. It does not correspond to the autoionization constant of water at this standard temperature.
12.
As temperature increases, the pH of pure water:
D.
Decreases below 7 but water remains neutral
ANSWER :
D. Decreases below 7 but water remains neutral
Approach
The question asks how the pH of pure water changes with increasing temperature. This involves understanding the autoionization of water and its temperature dependence, as well as the definition of pH and neutrality.
Step-by-step
- Autoionization of Water: Pure water undergoes a self-ionization process (autoionization) where water molecules react to form hydronium ions (\(\text{H}_3\text{O}^+\) or simply \(\text{H}^+\)) and hydroxide ions (\(\text{OH}^-\)).
\[ 2\text{H}_2\text{O}(l)
ightleftharpoons \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq) \] - Effect of Temperature on Autoionization: The autoionization of water is an endothermic process, meaning it absorbs heat. According to Le Chatelier's principle, if the temperature of an endothermic reaction is increased, the equilibrium will shift to the right (towards products) to absorb the added heat. This leads to an increase in the concentrations of both \([\text{H}^+]\) and \([\text{OH}^-]\).
- Ion Product of Water (\(K_w\)): The equilibrium constant for the autoionization of water is \(K_w = [\text{H}^+][\text{OH}^-]\). At \(25^\circ\text{C}\), \(K_w = 1.0 \times 10^{-14}\). Since increasing temperature shifts the equilibrium to the right, \(K_w\) increases with temperature. For example, at \(100^\circ\text{C}\), \(K_w \approx 1.0 \times 10^{-12}\).
- pH Calculation: In pure water, \([\text{H}^+] = [\text{OH}^-]\). Therefore, \([\text{H}^+] = \sqrt{K_w}\). The pH is defined as \(-\log_{10}[\text{H}^+]\).
At \(25^\circ\text{C}\): \([\text{H}^+] = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7} \text{ M}\). So, \(\text{pH} = -\log_{10}(1.0 \times 10^{-7}) = 7.0\).
At \(100^\circ\text{C}\): \([\text{H}^+] = \sqrt{1.0 \times 10^{-12}} = 1.0 \times 10^{-6} \text{ M}\). So, \(\text{pH} = -\log_{10}(1.0 \times 10^{-6}) = 6.0\). - Neutrality: Pure water is always neutral because the concentration of \(\text{H}^+\) ions is equal to the concentration of \(\text{OH}^-\) ions, regardless of the temperature. The definition of neutrality is \([\text{H}^+] = [\text{OH}^-]\), not necessarily \(\text{pH} = 7\). The \(\text{pH} = 7\) definition of neutrality is only valid at \(25^\circ\text{C}\).
- Conclusion: As temperature increases, \([\text{H}^+]\) increases (e.g., from \(10^{-7}\) to \(10^{-6}\)), which means the pH value decreases (e.g., from 7 to 6). However, because \([\text{H}^+]\) still equals \([\text{OH}^-]\), the water remains neutral.
Correct Option: D) Decreases below 7 but water remains neutral
Incorrect Options
- A) Becomes acidic — This is incorrect. Although the pH value decreases, the water does not become acidic. Acidity is defined by \([\text{H}^+] > [\text{OH}^-]\). In pure water, \([\text{H}^+] = [\text{OH}^-]\) always, so it remains neutral.
- B) Remains exactly 7 — This is incorrect. The pH of pure water is 7 only at \(25^\circ\text{C}\). As temperature increases, \(K_w\) increases, and thus \([\text{H}^+]\) increases, leading to a decrease in pH.
- C) Increases above 7 — This is incorrect. An increase in temperature shifts the autoionization equilibrium to produce more \(\text{H}^+\) ions, which means \([\text{H}^+]\) increases, and consequently, pH decreases (becomes less than 7). An increase above 7 would imply the water becomes basic.
