a) To calculate the pH value for \( [H^+] = 0.0003 \):
Method 1 (Formula):
Use the pH formula:
\( pH = -\log_{10}(0.0003) \)

Calculate:
\( pH \approx 3.522878745 \)

Round to two decimal places:
\( pH \approx 3.52 \)

Method 2 (Calculator):
Compute using a calculator:
\( -\log_{10}(0.0003) \approx 3.522878745 \)
Round to two decimal places:
\( pH \approx 3.52 \)

Thus:
The pH value is \( 3.52 \).

b) To calculate the hydrogen ion concentration in milk (\( pH = 6.6 \)):
Method 1 (Formula):
Use the inverse logarithm:
\( [H^+] = 10^{-6.6} \)

Calculate:
\( [H^+] \approx 2.51188643151 \times 10^{-7} \)

Thus:
The concentration is \( 2.51 \times 10^{-7} \) moles per litre.

Method 2 (Calculator):
Compute using a calculator:
\( 10^{-6.6} \approx 2.51188643151 \times 10^{-7} \)

Thus:
The concentration is \( 2.51 \times 10^{-7} \) moles per litre.

c) To calculate how many times stronger the acid in a lemon (\( pH = 2 \)) is compared to a tomato (\( pH = 4.5 \)):
Method 1 (Formula):
Calculate hydrogen ion concentrations:
\( [H^+]_{\text{lemon}} = 10^{-2} = 0.01 \)
\( [H^+]_{\text{tomato}} = 10^{-4.5} \approx 0.0000316227 \)

Find the ratio:
\( \frac{[H^+]_{\text{lemon}}}{[H^+]_{\text{tomato}}} = \frac{10^{-2}}{10^{-4.5}} = 10^{4.5 – 2} = 10^{2.5} \)
\( 10^{2.5} \approx 316.227766017 \)

Round to three significant figures:
\( \approx 316 \)

Method 2 (Calculator):
Compute concentrations:
\( [H^+]_{\text{lemon}} = 0.01 \)
\( [H^+]_{\text{tomato}} \approx 0.0000316227 \)

Calculate ratio:
\( \frac{0.01}{0.0000316227} \approx 316.227766017 \)

Round to three significant figures:
\( \approx 316 \)

Thus:
The acid in a lemon is approximately \( 316 \) times stronger than the acid in a tomato.