Answered

7.5 An impure sample of potassium hydroxide pellets with a mass of [tex]$11.2 \, \text{g}$[/tex] is added to [tex]0.09 \, \text{mole}[/tex] of sulphuric acid with a volume of [tex]600 \, \text{cm}^3[/tex]. It reacts according to the balanced chemical equation given below:

[tex]\[
H_2SO_4 (aq) + 2KOH (s) \rightarrow K_2SO_4 (aq) + 2H_2O (\ell)
\][/tex]

7.5.1 Calculate the initial pH of the sulphuric acid used in this reaction.

7.5.2 The percentage purity of the potassium hydroxide pellets used is [tex]80\%[/tex]. Calculate the number of moles of pure potassium hydroxide that react with [tex]H_2SO_4[/tex].



Answer :

GinnyAnswer
Let's tackle each part of the problem step-by-step:

### 7.5.1 Calculate the initial pH of the sulfuric acid used in this reaction.

Given Data:
- Volume of [tex]\( H_2SO_4 \)[/tex]: 600 cm[tex]\(^3\)[/tex] (or 0.600 L)
- Moles of [tex]\( H_2SO_4 \)[/tex]: 0.09 moles

#### Steps to Find Initial pH:

1. Find the concentration of sulfuric acid [tex]\( [H_2SO_4] \)[/tex]:

[tex]\[ \text{Concentration} = \frac{\text{Moles of } H_2SO_4}{\text{Volume (in liters)}} = \frac{0.09 \text{ moles}}{0.600 \text{ L}} \][/tex]

[tex]\[ \text{Concentration} = 0.15 \text{ M} \][/tex]

2. Since [tex]\( H_2SO_4 \)[/tex] dissociates completely in water, it produces [tex]\( 2 \)[/tex] moles of [tex]\( H^+ \)[/tex] ions per mole of [tex]\( H_2SO_4 \)[/tex]:

[tex]\[ [H^+] = 2 \times [H_2SO_4] \][/tex]

3. Calculate the concentration of [tex]\( H^+ \)[/tex] ions:

[tex]\[ [H^+] = 2 \times 0.15 = 0.30 \text{ M} \][/tex]

4. Calculate the initial pH:

[tex]\[ \text{pH} = -\log_{10} [H^+] \][/tex]

[tex]\[ \text{pH} = -\log_{10}(0.30) \][/tex]

The initial pH of the sulfuric acid is approximately:

[tex]\[ \text{pH} \approx 0.824 \][/tex]

### 7.5.2 Calculate the number of moles of pure potassium hydroxide that react with [tex]\( H_2SO_4 \)[/tex].

Given Data:
- Mass of impure [tex]\( KOH \)[/tex]: 11.2 g
- Purity: 80%
- Molar mass of [tex]\( KOH \)[/tex]: 56.11 g/mol

#### Steps to Find Moles of Pure Potassium Hydroxide:

1. Find the mass of pure [tex]\( KOH \)[/tex]:

[tex]\[ \text{Mass of pure } KOH = \left(\frac{\text{Purity}}{100}\right) \times \text{Mass of impure } KOH \][/tex]

[tex]\[ \text{Mass of pure } KOH = \left(\frac{80}{100}\right) \times 11.2 \text{ g} \][/tex]

[tex]\[ \text{Mass of pure } KOH = 8.96 \text{ g} \][/tex]

2. Convert the mass of pure [tex]\( KOH \)[/tex] to moles:

[tex]\[ \text{Moles of pure } KOH = \frac{\text{Mass of pure } KOH}{\text{Molar mass of } KOH} \][/tex]

[tex]\[ \text{Moles of pure } KOH = \frac{8.96 \text{ g}}{56.11 \text{ g/mol}} \][/tex]

The number of moles of pure potassium hydroxide is approximately:

[tex]\[ \text{Moles of pure } KOH \approx 0.160 \text{ moles} \][/tex]

### Summary:
- Initial pH of the sulfuric acid: [tex]\(\approx 0.824\)[/tex]
- Moles of pure potassium hydroxide: [tex]\(\approx 0.160\)[/tex] moles