Calculate the pH at 298 K of 5.0x10^-8 M NaOH, including the contribution of water (K_w = 1.0x10^-14). Give two decimals.

JEE Advanced levelNumerical answerDifficulty 3 of 3ie035

Calculate the pH at 298 K of $5.0 \times 1 0^{- 8} M$ $NaOH$ , including the contribution of water ( $K_{w} = 1.0 \times 1 0^{- 14}$ ). Give two decimals.

Numerical answer type, tolerance 0.01.

Total

[OH X^{-}] = C + [H X^{+}]

with

[H X^{+}] = K_{w} / [OH X^{-}]

, so

[OH X^{-}]^{2} - 5 \times 1 0^{- 8} [OH X^{-}] - 1 0^{- 14} = 0

and

[OH X^{-}] = 1.28 \times 1 0^{- 7} M

. Then

pOH = 6.893

and

pH = 7.107 \approx 7.11