To find the initial pH of the sulfuric acid before the reaction with potassium hydroxide, we need to follow a series of steps. Here’s a detailed, step-by-step solution:
### Step 1: Determine the volume of the sulfuric acid solution in liters
The volume of sulfuric acid given is 600 cm³. We need to convert this volume into liters, knowing that 1 liter = 1000 cm³.
[tex]\[
\text{Volume of } H_2SO_4 = \frac{600 \, \text{cm}^3}{1000 \, \text{cm}^3/\text{L}} = 0.6 \, \text{L}
\][/tex]
### Step 2: Calculate the concentration of sulfuric acid
The number of moles of sulfuric acid is given as 0.09 moles. To find the concentration (Molarity, M) of the sulfuric acid, we use the formula:
[tex]\[
\text{Concentration of } H_2SO_4 (\text{M}) = \frac{\text{moles of } H_2SO_4}{\text{volume of solution in L}}
\][/tex]
[tex]\[
\text{Concentration of } H_2SO_4 = \frac{0.09 \, \text{moles}}{0.6 \, \text{L}} = 0.15 \, \text{M}
\][/tex]
### Step 3: Determine the concentration of hydrogen ions
Sulfuric acid ([tex]\(H_2SO_4\)[/tex]) is a strong acid and dissociates completely in water. According to the dissociation equation:
[tex]\[
H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}
\][/tex]
Each molecule of [tex]\(H_2SO_4\)[/tex] produces 2 hydrogen ions ([tex]\(H^+\)[/tex]). Therefore, the concentration of hydrogen ions will be twice the concentration of [tex]\(H_2SO_4\)[/tex]:
[tex]\[
[\text{H}^+] = 2 \times \text{Concentration of } H_2SO_4
\][/tex]
[tex]\[
[\text{H}^+] = 2 \times 0.15 \, \text{M} = 0.3 \, \text{M}
\][/tex]
### Step 4: Calculate the initial pH
The pH of a solution is calculated using the formula:
[tex]\[
\text{pH} = -\log_{10} [\text{H}^+]
\][/tex]
Using the concentration of hydrogen ions:
[tex]\[
\text{pH} = -\log_{10} (0.3)
\][/tex]
Using the result directly from the calculations:
[tex]\[
\text{pH} \approx 0.52
\][/tex]
Therefore, the initial pH of the sulfuric acid used in this reaction is approximately [tex]\(0.52\)[/tex].
