Find the pH of $4.0 \times 1 0^{- 7} M$ $NaOH$ at 298 K, taking the ionisation of water into account (solve the resulting quadratic exactly). Give the answer to two decimal places.

Numerical answer type, tolerance 0.01.

Answer: 7.63

Total

[OH X^{-}] = 4 \times 1 0^{- 7} + y

with

[H X^{+}] = y

, and

y (4 \times 1 0^{- 7} + y) = 1 0^{- 14}

gives

y = 2.361 e - 08

[OH X^{-}] = 4.236 e - 07 M

pOH = 6.373

, so

pH = 7.63