Answer: 0.7
Before dilution; $[H^+] = 10^{-6} M$
$\therefore [H^+] from acid = 10^{-6} - 10^{-7} = 0.9 \times 10^{-6} M$
We Know that, $M_1V_1 = M_2V_2$
$\Rightarrow 0.9 \times 10^{-6} \times 100 = 1000 \times M_2$
$\Rightarrow M_2 = 0.9 \times 10^{-7} M$
After dilution; Total $[H^+] = [H^+]$ from acid + $[H^+]$ from water = $0.9 \times 10^{-7} M + 10^{-7} M = 1.9 \times 10^{-7} M$
$pH = - log [H^+] = - log 1.9 \times 10^{-7}$ = 7 - log 1.9 = 7 - 0.3 = 6.7
$\therefore$ Increase in pH = 6.7 - 6 = 0.7