Two solutions, A and B, each of 100L was made by dissolving 4g of...

{M}_{{H}_{2}S{O}_{4}} ⇒ \dfrac{9.8}{98 \times 100} = {10}^{- 3}$<br />${M}_{NaOH} ⇒ \dfrac{4}{40 \times 100} = {10}^{- 3}$<br />$= \dfrac{40 \times {10}^{- 3} - 10 \times {10}^{- 3} \times 2}{50} = \dfrac{20}{50} \times {10}^{- 3}$<br />\left[O{H}^{-}\right] = \dfrac{2}{5} \times {10}^{- 3} = 4 \times {10}^{- 4}$
$pOH = 4 - \log 4 = 3 . 4pH = 14 - 3 . 4 = 10 . 6$