The pH of a solution is given by

\mathrm{pH} = -\log\left(\frac{(\mathrm{H}^+)}{A}\right)

where (\mathrm{H}^+) is the number of hydrogen ions per liter and A is a constant.

a^x = b \iff x = \log_a b \,\,\,\,\,\left\lbrace\text{SL 1.5}\right\rbrace

The lemon juice has pH 3, so in 1 L:

-\log\left(\frac{(\mathrm{H}^+){\text{lemon}}}{A}\right) = 3

\frac{(\mathrm{H}^+){\text{lemon}}}{A} = 10^{-3}

(\mathrm{H}^+)_{\text{lemon}} = A \times 10^{-3}

In 30 mL (0.030\,\mathrm{L}) of lemon juice, the number of hydrogen ions is:

(\mathrm{H}^+)_{\text{lemon,\,30\,mL}} = A \times 10^{-3} \times 0.030 = 6.02 \times 10^{23} \times 10^{-3} \times 0.030 = 1.806 \times 10^{19}

The 330 mL (0.330\,\mathrm{L}) of pure water contains 1.987 \times 10^{16} hydrogen ions .

The total number of hydrogen ions after mixing is:

N_{\text{tot}} = 1.806 \times 10^{19} + 1.987 \times 10^{16} \approx 1.808 \times 10^{19}

The total volume after mixing is:

0.030 + 0.330 = 0.360\,\mathrm{L}

So, the concentration of hydrogen ions in the mixture is:

\left[\mathrm{H}^+\right] = \frac{N_{\text{tot}}}{0.360} = \frac{1.808 \times 10^{19}}{0.360} = 5.02 \times 10^{19}\,\text{ions\,L}^{-1}

Now, calculate the pH of the mixture:

\mathrm{pH} = -\log\left(\frac{5.02 \times 10^{19}}{6.02 \times 10^{23}}\right) = -\log(8.34 \times 10^{-5}) = 4.08

Problem Bank - Exponents & Logarithms