Two solutions, A and B, each of 100L was made by dissolving 4g of NaOH and 9.8g of H$$_2$$SO$$_4$$ in water, respectively. The pH of the resultant solution obtained from mixing 40L of solution A and 10L of solution B is (log 2 = 0.3)

We have solution A prepared by dissolving 4 g of NaOH in 100 L of water. The molar mass of NaOH is $$23 + 16 + 1 = 40\ \text{g mol}^{-1}.$$ Hence, the moles of NaOH present in solution A are

$$n_{\text{NaOH}} = \frac{4\ \text{g}}{40\ \text{g mol}^{-1}} = 0.1\ \text{mol}.$$

The volume is 100 L, so the molarity of NaOH in solution A is

$$[\text{NaOH}]_A = \frac{0.1\ \text{mol}}{100\ \text{L}} = 0.001\ \text{M}.$$

Next, solution B is prepared by dissolving 9.8 g of H$$_2$$SO$$_4$$ in 100 L of water. The molar mass of H$$_2$$SO$$_4$$ is $$2(1) + 32 + 4(16) = 98\ \text{g mol}^{-1}.$$ Thus, the moles of acid in solution B are

$$n_{\text{H}_2\text{SO}_4} = \frac{9.8\ \text{g}}{98\ \text{g mol}^{-1}} = 0.1\ \text{mol}.$$

Therefore, the molarity of H$$_2$$SO$$_4$$ in solution B is

$$[\text{H}_2\text{SO}_4]_B = \frac{0.1\ \text{mol}}{100\ \text{L}} = 0.001\ \text{M}.$$

Now we mix 40 L of solution A with 10 L of solution B.

Moles of NaOH taken from solution A:

$$n_{\text{NaOH,mix}} = 0.001\ \text{M} \times 40\ \text{L} = 0.04\ \text{mol}.$$

Moles of H$$_2$$SO$$_4$$ taken from solution B:

$$n_{\text{H}_2\text{SO}_4,\text{mix}} = 0.001\ \text{M} \times 10\ \text{L} = 0.01\ \text{mol}.$$

The neutralisation reaction is $$2\ \text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\ \text{H}_2\text{O}.$$ From the stoichiometry, 1 mol of H$$_2$$SO$$_4$$ requires 2 mol of NaOH.

NaOH required to neutralise 0.01 mol of H$$_2$$SO$$_4$$:

$$n_{\text{NaOH,needed}} = 2 \times 0.01 = 0.02\ \text{mol}.$$

We actually have 0.04 mol of NaOH, so the excess (unreacted) NaOH is

$$n_{\text{NaOH,excess}} = 0.04 - 0.02 = 0.02\ \text{mol}.$$

The total volume after mixing is

$$V_{\text{total}} = 40\ \text{L} + 10\ \text{L} = 50\ \text{L}.$$

Hence, the concentration of excess OH$$^-$$ ions is

$$[\text{OH}^-] = \frac{0.02\ \text{mol}}{50\ \text{L}} = 0.0004\ \text{M} = 4 \times 10^{-4}\ \text{M}.$$

Using the definition $$\text{pOH} = -\log[\text{OH}^-],$$ we get

$$\text{pOH} = -\log(4 \times 10^{-4}).$$

First, write $$\log(4 \times 10^{-4}) = \log 4 + \log 10^{-4}.$$ We know $$\log 10^{-4} = -4$$ and $$\log 4 = \log(2^2) = 2\log 2 = 2(0.3) = 0.6.$$ So,

$$\log(4 \times 10^{-4}) = 0.6 - 4 = -3.4.$$

Therefore, $$\text{pOH} = -(-3.4) = 3.4.$$

For water at 25 °C, $$\text{pH} + \text{pOH} = 14.$$ Substituting the pOH value,

$$\text{pH} = 14 - 3.4 = 10.6.$$

So, the answer is $$10.6$$.