Answer
a. $10^{-5.6}$ mole per liter.
b. $10^{-2.4}$ mole per liter.
c. The hydrogen ion concentration in part (b) is $10^{3.2}$ times greater than the one in part (a)
Work Step by Step
$\mathrm{a}$.
Insert $\mathrm{p}\mathrm{H}=5.6$ into the formula. Solve for x.
$5.6=-\log x$
$-5.6=\log x\qquad$....apply $10^{(..)}$ to both sides
$x=10^{-5.6}$ (leave as power of 10)
The hydrogen ion concentration is $10^{-5.6}$ mole per liter.
$\mathrm{b}$.
Insert $\mathrm{p}\mathrm{H}=2.4$ into the formula. Solve for x.
$2.4=-\log x$
$-2.4=\log x\qquad$....apply $10^{(..)}$ to both sides
$x=10^{-2.4}$ (leave as power of 10)
The hydrogen ion concentration is $10^{-2.4}$ mole per liter.
$\mathrm{c}$.
The ratio of the above results ( b to a):
(use $10^{m}\div 10^{n}=10^{m-n})$
$\displaystyle \frac{10^{-2.4}}{10^{-5.6}}=10^{-2.4-(-5.6)}=10^{3.2}$
The hydrogen ion concentration of the acidic rainfall in part (b) is $10^{3.2}$ times greater than the normal rainfall in part (a).
