Question 55 of 69intermediate🔧 ApplyNumerical3 marks

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?

Correct Answer

([H3O+]) 2×10−132 \times 10^{-13} M

([K+]) 0.05 M

([OH-]) 0.05 M

(pH) 12.70

Exercise: EXERCISES | Q: 6.57 | (Page 46)
For More Understanding

Explanation

To find the concentrations of the ions and the pH, we first determine the molarity of the KOH solution. Since KOH is a strong base, it dissociates completely to give equal concentrations of potassium and hydroxyl ions. The hydrogen ion concentration is then found using the ionic product of water, KwK_w, and the pH is calculated using the relation pKw=pH+pOH=14pK_w = \text{pH} + \text{pOH} = 14.

Solution Steps

  1. Step 1: Calculate the molarity of the KOH solution Molar mass of KOH = 39+16+1=56.139 + 16 + 1 = 56.1 g/mol Moles of KOH = 0.561 g56.1 g/mol=0.01\frac{0.561 \text{ g}}{56.1 \text{ g/mol}} = 0.01 mol Volume of solution = 200 mL=0.2 L200 \text{ mL} = 0.2 \text{ L} Molarity of KOH = 0.01 mol0.2 L=0.05 M\frac{0.01 \text{ mol}}{0.2 \text{ L}} = 0.05 \text{ M}

  2. Step 2: Calculate concentrations of potassium and hydroxyl ions KOH dissociates completely in water:

    KOH→K++OH−\text{KOH} \rightarrow \text{K}^+ + \text{OH}^-

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

  3. Step 3: Calculate the concentration of hydrogen (hydronium) ions At 298 K, Kw=[H3O+][OH−]=10−14K_w = [\text{H}_3\text{O}^+] [\text{OH}^-] = 10^{-14}

    [H3O+]=10−14[OH−]=10−140.05=10−145×10−2=2×10−13 M[\text{H}_3\text{O}^+] = \frac{10^{-14}}{[\text{OH}^-]} = \frac{10^{-14}}{0.05} = \frac{10^{-14}}{5 \times 10^{-2}} = 2 \times 10^{-13} \text{ M}

  4. Step 4: Calculate the pH of the solution

    pOH=−log⁡[OH−]=−log⁡(0.05)=−log⁡(5×10−2)=2−0.6990=1.3010\text{pOH} = -\log [\text{OH}^-] = -\log (0.05) = -\log (5 \times 10^{-2}) = 2 - 0.6990 = 1.3010

    Using pKw=pH+pOH=14pK_w = \text{pH} + \text{pOH} = 14:

    pH=14−pOH=14−1.3010=12.699≈12.70\text{pH} = 14 - \text{pOH} = 14 - 1.3010 = 12.699 \approx 12.70
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