Question #8818a

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 #"pH"# equal to #0.7#, so you can say that the concentration of hydronium cations will be equal to

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

This means that the #"14-L"# solution will contain

#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 #"pH"# equal to #2.3#, which implies that its concentration of hydronium cations must be

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

If you take #V# #"L"# to be the total volume of the target solution, you can say that

#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 #"pH"# of the solution from #0.7# to #2.3#, the volume of the solution must increase from #"14 L"# to #"557.4 L"#.

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 #"pH"# means one 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 #"557.4 L"#, but don't forget to round the final answer to 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.