# Question a1253

Nov 13, 2016

Here's what I got.

#### Explanation:

The concentration of hydronium cations, ${\text{H"_3"O}}^{+}$, referred to here as hydrogen ions, ${\text{H}}^{+}$, is related to the solution's pH by the equation

color(blue)(ul(color(black)("pH" = - log(["H"^(+)])))

To find the concentration of hydrogen ions, rearrange the above equation as

log(["H"^(+)]) = - "pH"

This can be written by using powers of $10$

10^(log(["H"^(+)])) = 10^(-"pH")

which is equivalent to

$\left[\text{H"^(+)] = 10^(-"pH}\right)$

Plug in your value to find

color(darkgreen)(ul(color(black)(["H"^(+)] = 10^(-2.89) = 1.3 * 10^(-3)"M")))#

The answer is rounded to two sig figs, the number of decimal places you have for the pH of the solution, because it does not depend on the volume of the sample.

The number of moles of hydrogen ions, however, does depend on the volume of the sample, so it will be rounded to one sig fig.

Now, molarity is defined as the number of moles of solute present in one liter of solution. In your case, a $1.3 \cdot {10}^{- 3} \text{-M}$ solution will contain $1.3 \cdot {10}^{- 3}$ moles of hydrogen ions for every liter of solution.

As you know, one liter is equivalent to

${\text{1 L" = "1 dm"^3 = 10^3"cm}}^{3}$

This means that your sample contains

$30 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{cm"^3))) * (1.3 * 10^(-3)"moles H"^(+))/(10^3color(red)(cancel(color(black)("cm"^3)))) = color(darkgreen)(ul(color(black)(4 * 10^(-5)"moles H}}^{+}}}}$