Skip to content
Exercises · 6.57

Q.If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?

Jammu Kashmir JkboseTextbookSubjective· 3mImportance★★★★★est
55% · 85/155 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Dissolving KOH in water produces a strong base that dissociates completely. The concentration of OH−\text{OH}^- is found from molarity, then [H+][\text{H}^+] follows from KwK_w, and pH from −log⁡[H+]-\log[\text{H}^+]. Final answer: [K+]=[OH−]=0.05 M[\text{K}^+] = [\text{OH}^-] = 0.05 \, \text{M}, [H+]=2×10−13 M[\text{H}^+] = 2 \times 10^{-13} \, \text{M}, pH =12.70= 12.70.

Potassium hydroxide is a strong base, meaning it dissociates completely in water. When KOH dissolves, every formula unit splits into one potassium ion and one hydroxide ion:

KOH⟶K++OH−\text{KOH} \longrightarrow \text{K}^+ + \text{OH}^-

Because the dissociation is complete, the concentration of K+\text{K}^+ equals the concentration of OH−\text{OH}^-, and both equal the molarity of the KOH solution. Once we know [OH−][\text{OH}^-], we can find [H+][\text{H}^+] using the water equilibrium constant Kw=[H+][OH−]=1.0×10−14K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} at 298 K. The pH then follows directly from the hydrogen ion concentration.

Step-by-step calculation

1. Find the number of moles of KOH

The molar mass of KOH is:

MKOH=39 (K)+16 (O)+1 (H)=56 g/molM_{\text{KOH}} = 39 \, (\text{K}) + 16 \, (\text{O}) + 1 \, (\text{H}) = 56 \, \text{g/mol}

Number of moles:

n=0.561 g56 g/mol=0.01 moln = \frac{0.561 \, \text{g}}{56 \, \text{g/mol}} = 0.01 \, \text{mol}

2. Calculate the molarity of the KOH solution

Volume in litres is 200 mL=0.200 L200 \, \text{mL} = 0.200 \, \text{L}.

[KOH]=0.01 mol0.200 L=0.05 M[\text{KOH}] = \frac{0.01 \, \text{mol}}{0.200 \, \text{L}} = 0.05 \, \text{M}

3. Determine the concentration of potassium and hydroxyl ions

Since KOH dissociates completely:

[K+]=[OH−]=0.05 M[\text{K}^+] = [\text{OH}^-] = 0.05 \, \text{M}

[K+]=[OH−]=0.05 M[\text{K}^+] = [\text{OH}^-] = 0.05 \, \text{M}

4. Calculate the hydrogen ion concentration

Using the ionic product of water at 298 K:

Kw=[H+][OH−]=1.0×10−14K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} …

Start your 14-day free trial to unlock the full solution →

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests
Start 14-day free trial →See Plans & Pricing →

Full access for 14 days. No credit card required.

←Previous question · EXERCISES6.56 — The pH of milk, black coffee, tomato juice, lemon juice and egg white …→Next question · EXERCISES6.58 — The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate…