Explanation

(b)
$\left\lbrack OH^{-}\ \right\rbrack in\ the\ diluted\ base = \frac{10^{- 6}}{10^{2}} = 10^{- 8}$
Total $\left\lbrack OH^{-} \right\rbrack = 10^{- 8} + \left\lbrack OH^{-} \right\rbrack\ of\ water$
$= (10^{- 8} + 10^{- 7})$M
$= 10^{- 8}(1 + 10)$M
$= 11 \times 10^{- 8}$ M
$pOH = - \log{11 \times 10^{- 8}}$
$= - \log{11 + 8\log 10}$
$= 6.9586$
$pH = 14 - 6.9586$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 7.0414$