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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?

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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} …

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