Answer
a. $10^{-2.3}$ mole per liter.
b. $10^{-1}$ mole per liter.
c. The hydrogen ion concentration in part (b) is $10^{1.3}$ times greater than the concentration in part (a)
Work Step by Step
$\mathrm{a}$.
Insert $\mathrm{p}\mathrm{H}=2.3$ into the formula. Solve for x.
$2.3=-\log x$
$-2.3=\log x\qquad$....apply $10^{(..)}$ to both sides
$x=10^{-2.3}$ (leave as power of 10)
The hydrogen ion concentration is $10^{-2.3}$ mole per liter.
$\mathrm{b}$.
Insert $\mathrm{p}\mathrm{H}=$1 into the formula. Solve for x.
$1=-\log x$
$-1=\log x\qquad$....apply $10^{(..)}$ to both sides
$x=10^{-1}$ (leave as power of 10)
The hydrogen ion concentration is $10^{-1}$ mole per liter.
$\mathrm{c}$.
The ratio of the above results ( b to a):
(use $10^{m}\div 10^{n}=10^{m-n})$
$\displaystyle \frac{10^{-1}}{10^{-2.3}}=10^{-1-(-2.3)}=10^{1.3}$
The concentration in part (b) is $10^{1.3}$ times greater than the concentration in part (a).
