# Question #8818a

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

The idea here is that you're required to calculate the volume of water that would be needed in order to **dilute** the hydrochloric acid solution, i.e. to **decrease** the concentration of hydronium cations.

As you know, the concentration of hydronium cations is given by

#color(blue)(ul(color(black)(["H"_3"O"^(+)] = 10^(-"pH"))))#

The initial solution has a

#["H"_ 3"O"^(+)]_ 0 = 10^(-0.7) quad "M"#

This means that the

#14 color(red)(cancel(color(black)("L solution"))) * (10^(-0.7) quad "moles H"_3"O"^(+))/(1color(red)(cancel(color(black)("L solution")))) = "2.7934 moles H"_3"O"^(+)#

Now, when you dilute this solution, the number of moles of hydronium cations remains unchanged.

The target solution must have a

#["H"_ 3"O"^(+)]_ "fin" = 10^(-2.3) quad "M"#

If you take

#V = (2.7934 color(red)(cancel(color(black)("moles H"_3"O"^(+)))))/(10^(-2.3) color(red)(cancel(color(black)("moles H"_3"O"^(+)))) "L"^(-1)) = "557.4 L"#

So, you know that in order to **decrease** the **increase** from

This means that the initial solution **must be added to** a volume of water equal to--remember, you must **always** add strong acids to water, **never** the other way around!

#V_"water" = "557.4 L " - " 14 L" = color(darkgreen)(ul(color(black)("540 L")))#

I'll leave the answer rounded to two **sig figs**, no decimal places, but keep in mind that one **decimal place** for the **significant figure** for the concentration of hydronium cations.

So the **volume** of the target solution *should* be rounded to one significant figure

#"557.4 L " ~~ " 600 L"#

Now, because you have one more operation to perform, you can leave the volume of the solution as **one significant figure** and no decimal places.

Perform the subtraction normally

#"557.4 L " - " 14 L" = "543 L"#

and round the final answer to

#"543 L " ~~ " 500 L"#

This is actually a great example of how significant figures can affect the final answer.