Question #dcad3

1 Answer
Nathan L.
Sep 3, 2017

#["H"^+] = 1.3xx10^-8# #"mol"*"L"^-1#

Explanation:

We're asked to find the hydrogen ion concentration, #["H"^+]#, of a solution, given its #"pH"#.

To do this, we can use the following relationship that is worth knowing:

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

Which we can rearrange to solve for the hydrogen ion concentration:

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

We're given that the #"pH"# is #7.90#, so we have

#["H"^+] = 10^-7.90 = color(red)(ulbar(|stackrel(" ")(" "1.3xx10^-8color(white)(l)"mol"*"L"^-1" ")|)#

#" "#

Becuase the #"pH"# value was given to two decimal places, the concentration is expressed with two significant figures.

The reason behind this comes from the fact that only the decimal places are important when calculating the logarithm. Basically, if the #["H"^+]# was a very low value, such as #1.3xx10^-13# #"mol/L"# (which contains two significant figures also), then the #"pH"# of the solution would be

#"pH" = -log(1.3xx10^-13color(white)(l)"mol"*"L"^-1) = 13#

if we only used two significant figures. This value isn't very precise, so we just use two decimal places:

#"pH" = 12.87#

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