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A sulfuric acid solution is prepared by dissolving [tex]7.35 \, \text{g}[/tex] of [tex]H_2SO_4 (l)[/tex] in [tex]500 \, \text{cm}^3[/tex] of water.

7.2.1 Calculate the number of moles of [tex]H_2SO_4[/tex] present in this solution.

Sodium hydroxide [tex](NaOH)[/tex] pellets are added to the [tex]500 \, \text{cm}^3[/tex] [tex]H_2SO_4[/tex] solution. The balanced equation for the reaction is:

[tex]
H_2SO_4 (aq) + 2 NaOH (s) \rightarrow Na_2SO_4 (aq) + 2 H_2O (l)
[/tex]

After completion of the reaction, the pH of the solution was found to be 1.3. Assume complete ionization of [tex]H_2SO_4[/tex].

7.2.2 Calculate the mass of NaOH added to the [tex]H_2SO_4[/tex] solution. Assume that the volume of the solution does not change.



Answer :

GinnyAnswer
Let's break down the solution step by step.

### Part 7.2.1: Calculate the number of moles of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] present in this solution

1. Given data:
- Mass of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] = 7.35 g
- Molar mass of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] = 98.079 g/mol

2. Formula to calculate moles:
[tex]\[ \text{Moles} = \frac{\text{mass}}{\text{molar mass}} \][/tex]

3. Calculation:
[tex]\[ \text{Moles of } \text{H}_2\text{SO}_4 = \frac{7.35 \text{ g}}{98.079 \text{ g/mol}} = 0.0749 \text{ moles} \][/tex]

### Part 7.2.2: Calculate the mass of [tex]\( \text{NaOH} \)[/tex] added to the [tex]\( \text{H}_2\text{SO}_4 \)[/tex] solution

1. Balanced chemical equation:
[tex]\[ \text{H}_2\text{SO}_4 (aq) + 2 \text{NaOH} (s) \rightarrow \text{Na}_2\text{SO}_4 (aq) + 2 \text{H}_2\text{O} (l) \][/tex]

2. Number of moles of [tex]\( \text{H}_2\text{SO}_4 \)[/tex]:
[tex]\[ \text{Moles of } \text{H}_2\text{SO}_4 = 0.0749 \text{ moles} \][/tex]

3. Since the reaction requires 2 moles of [tex]\( \text{NaOH} \)[/tex] for every 1 mole of [tex]\( \text{H}_2\text{SO}_4 \)[/tex]:
[tex]\[ \text{Moles of } \text{NaOH} \text{ needed} = 2 \times \text{Moles of } \text{H}_2\text{SO}_4 = 2 \times 0.0749 = 0.1499 \text{ moles} \][/tex]

4. Molar mass of [tex]\( \text{NaOH} \)[/tex]:
[tex]\[ \text{Molar mass of } \text{NaOH} = 39.997 \text{ g/mol} \][/tex]

5. Formula to calculate mass:
[tex]\[ \text{Mass} = \text{moles} \times \text{molar mass} \][/tex]

6. Calculation:
[tex]\[ \text{Mass of } \text{NaOH} = 0.1499 \text{ moles} \times 39.997 \text{ g/mol} = 5.995 \text{ grams} \][/tex]

Therefore:
- The number of moles of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] in the solution is [tex]\(0.0749\)[/tex] moles.
- The mass of [tex]\( \text{NaOH} \)[/tex] added to the [tex]\( \text{H}_2\text{SO}_4 \)[/tex] solution is [tex]\(5.995\)[/tex] grams.

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