Answer
from $1.58\times 10^{-4}$ M to $1.58\times 10^{-3}$ M
Work Step by Step
(see p.381)
The $\mathrm{p}\mathrm{H}$ scale measures the acidity of a solution:$ \quad \mathrm{p}\mathrm{H}=-\log[\mathrm{H}^{+}]$
where $[\mathrm{H}^{+}]$ is the hydrogen ion concentration (in moles per liter, M).
Solutions are
neutral if $pH=7$
acidic if $pH < 7$
basic if $pH > 7$
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Solving for $[\mathrm{H}^{+}]$,
$pH=-\log[\mathrm{H}^{+}]\qquad /\times(-1)$
$\log[\mathrm{H}^{+}]=-pH\qquad /$... apply $10^{(...)}$ to both sides
$[\mathrm{H}^{+}]=10^{-pH}$
$pH=2.8\Rightarrow[\mathrm{H}^{+}]=10^{-2.8}\approx 1.58\times 10^{-3}$ M
$pH=3.8\Rightarrow[\mathrm{H}^{+}]=10^{-3.8}\approx 1.58\times 10^{-4}$ M
The range of $[\mathrm{H}^{+}]$ is approximately
from $1.58\times 10^{-4}$ M to $1.58\times 10^{-3}$ M
