Acids, Bases, and Electrochemistry
Acids, Bases, and Electrochemistry MCQ Questions
1.
What is the pH of a 0.001 M HCl solution at 25°C?
Approach
To find the pH of an HCl solution, we need to determine the concentration of hydrogen ions (\([\text{H}^+]\)). Since HCl is a strong acid, it dissociates completely in water. The pH is then calculated using the formula \( \text{pH} = -\log_{10}[\text{H}^+] \).
Step-by-step
- Identify the nature of the acid: HCl (hydrochloric acid) is a strong acid. This means it dissociates completely in an aqueous solution.
- Write the dissociation equation: When HCl dissolves in water, it dissociates as follows:\[ \text{HCl} \text{(aq)} \rightarrow \text{H}^+ \text{(aq)} + \text{Cl}^- \text{(aq)} \]
- Determine the concentration of \([\text{H}^+]\): Since HCl dissociates completely, the concentration of \(\text{H}^+\) ions will be equal to the initial concentration of HCl.Given concentration of HCl \(= 0.001 \text{ M}\).Therefore, \([\text{H}^+] = 0.001 \text{ M}\).
- Express \([\text{H}^+]\) in scientific notation:\[ [\text{H}^+] = 0.001 \text{ M} = 1 \times 10^{-3} \text{ M} \]
- Calculate the pH: The pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration:\[ \text{pH} = -\log_{10}[\text{H}^+] \]Substitute the value of \([\text{H}^+]\):\[ \text{pH} = -\log_{10}(1 \times 10^{-3}) \]Using the logarithm property \(\log(a \times b) = \log a + \log b\) and \(\log(a^b) = b \log a\):\[ \text{pH} = -(\log_{10}1 + \log_{10}10^{-3}) \]Since \(\log_{10}1 = 0\) and \(\log_{10}10^{-3} = -3\):\[ \text{pH} = -(0 + (-3)) \]\[ \text{pH} = -(-3) \]\[ \text{pH} = 3 \]
Correct Option: D) 3
Incorrect Options
- A) 2 — This would be the pH if the \([\text{H}^+]\) concentration were \(1 \times 10^{-2} \text{ M}\) (i.e., 0.01 M HCl).
- B) 1 — This would be the pH if the \([\text{H}^+]\) concentration were \(1 \times 10^{-1} \text{ M}\) (i.e., 0.1 M HCl).
- C) 11 — This pH indicates a basic solution, which is incorrect for an HCl solution. A pH of 11 would correspond to a \([\text{H}^+]\) concentration of \(1 \times 10^{-11} \text{ M}\) or a \([\text{OH}^-]\) concentration of \(1 \times 10^{-3} \text{ M}\).
2.
The pH of a 0.01 M NaOH solution is:
Approach
To find the pH of a NaOH solution, we first determine the concentration of hydroxide ions (\([\text{OH}^-]\)). Since NaOH is a strong base, it dissociates completely in water. From \([\text{OH}^-]\), we can calculate pOH, and then use the relationship between pH and pOH to find the pH.
Step-by-step
- Identify the nature of the solution: NaOH is a strong base. This means it dissociates completely in water according to the reaction:\[ \text{NaOH} \text{(aq)} \rightarrow \text{Na}^+ \text{(aq)} + \text{OH}^- \text{(aq)} \]
- Determine the concentration of hydroxide ions (\([\text{OH}^-]\)): Since NaOH dissociates completely, the concentration of \(\text{OH}^-\) ions will be equal to the initial concentration of NaOH.Given concentration of NaOH = \(0.01 \text{ M}\).Therefore, \([\text{OH}^-]\) = \(0.01 \text{ M}\).This can be written as \(1 \times 10^{-2} \text{ M}\).
- Calculate the pOH: The pOH is defined as the negative logarithm (base 10) of the hydroxide ion concentration:\[ \text{pOH} = -\log_{10}[\text{OH}^-] \]Substitute the value of \([\text{OH}^-]\):\[ \text{pOH} = -\log_{10}(1 \times 10^{-2}) \]Using the logarithm property \(\log(a \times b) = \log a + \log b\) and \(\log(10^x) = x\):\[ \text{pOH} = -(\log_{10}1 + \log_{10}10^{-2}) \]\[ \text{pOH} = -(0 + (-2)) \]\[ \text{pOH} = 2 \]
- Calculate the pH: At \(25^\circ\text{C}\), the sum of pH and pOH is always 14:\[ \text{pH} + \text{pOH} = 14 \]Substitute the calculated pOH value:\[ \text{pH} + 2 = 14 \]\[ \text{pH} = 14 - 2 \]\[ \text{pH} = 12 \]
Correct Option: C) 12
Incorrect Options
- A) 14 — This would be the pH of a \(1 \text{ M}\) strong base solution, not \(0.01 \text{ M}\). If \([\text{OH}^-]\) were \(1 \text{ M}\), then \(\text{pOH} = -\log(1) = 0\), and \(\text{pH} = 14 - 0 = 14\).
- B) 10 — This pH would correspond to a pOH of 4, meaning \([\text{OH}^-]\) would be \(1 \times 10^{-4} \text{ M}\), which is incorrect for a \(0.01 \text{ M}\) NaOH solution.
- D) 2 — This is the pOH of the solution, not the pH. A pH of 2 would
3.
If the [H⁺] of a solution is 1 × 10⁻⁵ M, what is its pH?
Approach
The pH of a solution is a measure of its acidity or alkalinity and is defined by the negative logarithm (base 10) of the hydrogen ion concentration, \([\text{H}^+]\). We will use the formula for pH to calculate its value given the hydrogen ion concentration.
Step-by-step
- Recall the pH formula: The pH is defined as the negative base-10 logarithm of the hydrogen ion concentration, \([\text{H}^+]\).
\[ \text{pH} = -\log_{10}[\text{H}^+] \] - Substitute the given concentration: The problem states that the hydrogen ion concentration, \([\text{H}^+]\), is \(1 \times 10^{-5} \text{ M}\). Substitute this value into the pH formula.
\[ \text{pH} = -\log_{10}(1 \times 10^{-5}) \] - Simplify the logarithm: Using the logarithm property \(\log(a \times b) = \log(a) + \log(b)\) and \(\log(1) = 0\):
\[ \text{pH} = -(\log_{10}(1) + \log_{10}(10^{-5})) \]
\[ \text{pH} = -(0 + (-5)) \]
\[ \text{pH} = -(-5) \]
\[ \text{pH} = 5 \]
Thus, the pH of the solution is 5.
Correct Option:
D) 5 is the correct pH value calculated from the given hydrogen ion concentration.
Incorrect Options
- A) 9 — This value would correspond to a hydrogen ion concentration of \(1 \times 10^{-9} \text{ M}\), which is a basic solution. The given concentration indicates an acidic solution.
- B) 10⁻⁵ — This is the hydrogen ion concentration itself, not the pH. pH is a logarithmic scale derived from the concentration.
- C) 4 — This value would correspond to a hydrogen ion concentration of \(1 \times 10^{-4} \text{ M}\). While it's an acidic pH, it does not match the given concentration of \(1 \times 10^{-5} \text{ M}\).
4.
The pH of pure water at 25°C is exactly 7 because:
A.
Water contains no ions
B.
Water is neutral by definition
ANSWER :
C. [H⁺] = [OH⁻] = 10⁻⁷ M
Approach
The question asks why the pH of pure water at 25°C is exactly 7. This requires understanding the autoionization of water and the definition of pH.
Step-by-step
- Autoionization of Water: Pure water undergoes a process called autoionization (or autoprotolysis), where a small fraction of water molecules dissociate into hydrogen ions (H⁺, often represented as H₃O⁺) and hydroxide ions (OH⁻). The equilibrium reaction is:\[ \text{H}_2\text{O}
ightleftharpoons \text{H}^+ + \text{OH}^- \]
- Equilibrium Constant (Kw): The equilibrium constant for this reaction is called the ion product of water, \(K_w\). At 25°C, the value of \(K_w\) is \(1.0 \times 10^{-14}\).\[ K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} \text{ at } 25^\circ\text{C} \]
- Concentrations in Pure Water: In pure water, for every H⁺ ion formed, one OH⁻ ion is also formed. Therefore, the concentration of H⁺ ions must be equal to the concentration of OH⁻ ions.\[ [\text{H}^+] = [\text{OH}^-] \]
- Calculating Concentrations: Substituting this into the \(K_w\) expression:\[ [\text{H}^+]^2 = 1.0 \times 10^{-14} \]Taking the square root of both sides:\[ [\text{H}^+] = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7} \text{ M} \]Thus, \( [\text{H}^+] = [\text{OH}^-] = 1.0 \times 10^{-7} \text{ M} \) in pure water at 25°C.
- Calculating pH: The pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration:\[ \text{pH} = -\log_{10}[\text{H}^+] \]Substituting the calculated concentration:\[ \text{pH} = -\log_{10}(1.0 \times 10^{-7}) = -(-7) = 7 \]
Correct Option: C) [H⁺] = [OH⁻] = 10⁻⁷ M
This statement directly leads to the pH of 7. As derived above, in pure water at 25°C, the autoionization of water produces equal concentrations of H⁺ and OH⁻ ions, both of which are \(10^{-7}\) M. This concentration then results in a pH of 7.
Incorrect Options
- A) Water contains no ions — This is incorrect. Pure water, though a poor conductor, does contain a very small but equal concentration of H⁺ and OH⁻ ions due to autoionization. Without these ions, pH would not be defined.
- B) Water is neutral by definition — While water is neutral, this is a consequence of its properties, not the fundamental reason for its pH being exactly 7. The neutrality arises from the equal concentrations of H⁺ and OH⁻ ions, which is explained by option C.
- D) Kw = 10⁻⁷
5.
What is the pOH of a solution whose pH is 3.5?
Approach
The relationship between pH and pOH in an aqueous solution at 25°C is given by a fundamental equation. We will use this equation to calculate the pOH when the pH is known.
Step-by-step
- Recall the relationship between pH and pOH:
For any aqueous solution at 25°C, the sum of pH and pOH is always 14.
\[ \text{pH} + \text{pOH} = 14 \] - Substitute the given pH value into the equation:
The problem states that the pH of the solution is 3.5.
\[ 3.5 + \text{pOH} = 14 \] - Solve for pOH:
Subtract 3.5 from both sides of the equation to find the pOH.
\[ \text{pOH} = 14 - 3.5 \]\[ \text{pOH} = 10.5 \]
Correct Option:
D) 10.5 — The pOH of the solution is 10.5, calculated using the relationship \(\text{pH} + \text{pOH} = 14\).
Incorrect Options
- A) 3.5 — This is the given pH value, not the pOH.
- B) 7.0 — This would be the pH and pOH of a neutral solution. The given pH is acidic (3.5), so the pOH must be basic.
- C) 11.5 — This value would result if the pH was 2.5 (i.e., \(14 - 2.5 = 11.5\)), not 3.5.
6.
When the pH of a solution changes from 4 to 6, the H⁺ ion concentration:
ANSWER :
D. Decreases 100 times
Approach
The pH scale is a logarithmic scale used to specify the acidity or basicity of an aqueous solution. It is inversely related to the concentration of hydrogen ions (\(\text{H}^+\)). A change in pH by one unit represents a tenfold change in \(\text{H}^+\) ion concentration. We will use the definition of pH to calculate the change in \(\text{H}^+\) ion concentration.
Step-by-step
- Recall the definition of pH:
The pH of a solution is defined as the negative base-10 logarithm of the hydrogen ion concentration, expressed in moles per liter.
\[ \text{pH} = -\log_{10}[\text{H}^+] \] - Express \(\text{H}^+\) ion concentration in terms of pH:
From the definition, we can rearrange the formula to find the \(\text{H}^+\) ion concentration:
\[ [\text{H}^+] = 10^{-\text{pH}} \] - Calculate initial \(\text{H}^+\) ion concentration (at pH 4):
When \(\text{pH}_1 = 4\), the initial \(\text{H}^+\) ion concentration is:
\[ [\text{H}^+]_1 = 10^{-4} \text{ M} \] - Calculate final \(\text{H}^+\) ion concentration (at pH 6):
When \(\text{pH}_2 = 6\), the final \(\text{H}^+\) ion concentration is:
\[ [\text{H}^+]_2 = 10^{-6} \text{ M} \] - Determine the factor of change:
To find out how many times the \(\text{H}^+\) ion concentration has changed, we divide the initial concentration by the final concentration (or vice versa, depending on how the question is phrased; here we want to see if it increased or decreased and by what factor).
\[ \frac{[\text{H}^+]_1}{[\text{H}^+]_2} = \frac{10^{-4}}{10^{-6}} = 10^{(-4 - (-6))} = 10^{(-4 + 6)} = 10^2 = 100 \] - Interpret the result:
Since \([\text{H}^+]_1 = 100 \times [\text{H}^+]_2\), it means the initial \(\text{H}^+\) ion concentration was 100 times greater than the final \(\text{H}^+\) ion concentration. Therefore, when the pH changes from 4 to 6, the \(\text{H}^+\) ion concentration decreases 100 times.
Correct Option:
D) Decreases 100 times. As the pH increases from 4 to 6, the solution becomes less acidic, meaning the concentration of \(\text{H}^+\) ions decreases. A change of 2 pH units corresponds to a \(10^2 = 100\)-fold change in \(\text{H}^+\) ion concentration. Since pH increased, \(\text{H}^+\) concentration decreased.
Incorrect Options
- A) Increases 100 times — This would be true
