Answer
Decreases to $\displaystyle \frac{1}{10}$ of the initial amount.
Work Step by Step
Using the work of part b,
$H^{+}=10^{-p_{H}}$
If $p_{H}$ increases by 1,
$H^{+}=10^{-(p_{H}+1)}=10^{-p_{H}-1}=10^{-p_{H}}\times(10^{-1})$
The concentration has decreased by a factor of $10^{-1}=\displaystyle \frac{1}{10}$
or, decreases to $\displaystyle \frac{1}{10}$ of the initial amount.
