# Question #a1253

##### 1 Answer

Here's what I got.

#### Explanation:

The concentration of hydronium cations, *hydrogen ions*,

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

which is equivalent to

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

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 **moles** of hydrogen ions **for every liter** of solution.

As you know, one liter is equivalent to

#"1 L" = "1 dm"^3 = 10^3"cm"^3#

This means that your sample contains

#30 color(red)(cancel(color(black)("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"^(+))))#