Question:
\(2.8\) g of KOH is dissolved in a \(500\) mL solution at \(298\) K. What is the \(p^H\) of solution ? (molar mass of KOH = \(56\) g/mol)
\(2.8\) g of KOH is dissolved in a \(500\) mL solution at \(298\) K. What is the \(p^H\) of solution ? (molar mass of KOH = \(56\) g/mol)
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KOH is a strong base: find [OH-] from moles and volume, then pOH and pH = 14 - pOH.
Updated On: Oct 1, 2026
- \(1\)
- \(8\)
- \(11\)
- \(13\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
KOH is a strong base and dissociates completely, so \([\text{OH}^-]\) equals the molarity of KOH. At 298 K, \(\text{pH} + \text{pOH} = 14\).
Step 2: Find the moles and molarity:
\(n = \dfrac{2.8}{56} = 0.05\) mol. Volume is 500 mL = 0.5 L.
\[ M = \dfrac{0.05}{0.5} = 0.1 \text{ M} \]
Step 3: Find pOH and pH:
\([\text{OH}^-] = 0.1 = 10^{-1}\) M, so \(\text{pOH} = 1\).
\[ \text{pH} = 14 - 1 = 13 \]
Step 4: Why the other options are wrong.
A pH of 1 is what we get if we forget to convert pOH to pH. A pH of 8 or 11 would need much lower hydroxide concentration (\(10^{-6}\) M or \(10^{-3}\) M).
Final Answer:
The pH of the solution is 13. \[ \boxed{\text{(D) }13} \]
KOH is a strong base and dissociates completely, so \([\text{OH}^-]\) equals the molarity of KOH. At 298 K, \(\text{pH} + \text{pOH} = 14\).
Step 2: Find the moles and molarity:
\(n = \dfrac{2.8}{56} = 0.05\) mol. Volume is 500 mL = 0.5 L.
\[ M = \dfrac{0.05}{0.5} = 0.1 \text{ M} \]
Step 3: Find pOH and pH:
\([\text{OH}^-] = 0.1 = 10^{-1}\) M, so \(\text{pOH} = 1\).
\[ \text{pH} = 14 - 1 = 13 \]
Step 4: Why the other options are wrong.
A pH of 1 is what we get if we forget to convert pOH to pH. A pH of 8 or 11 would need much lower hydroxide concentration (\(10^{-6}\) M or \(10^{-3}\) M).
Final Answer:
The pH of the solution is 13. \[ \boxed{\text{(D) }13} \]
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