#30.0" cm"^3# sample of a #0.480" mol/L"# solution of potassium hydroxide was partially neutralized by the addition of #18.0" cm"^3# of a #0.350" mol/L"# solution of sulfuric acid. Calculate the pH of the solution formed?

For starters, I would say that if the values you have here are correct, then you got this one right.

Your goal here is to figure out what concentration of hydroxide anions remains after the incomplete neutralization of the potassium hydroxide solution.

You know that potassium hydroxide and sulfuric acid react in a #2:1# mole ratio to produce aqueous potassium sulfate and water.

#2"KOH"_ ((aq)) + "H"_ 2"SO"_ (4(aq)) -> "K"_ 2"SO"_ (4(aq)) + "H"_ 2"O"_ ((l))#

In your case, the two solutions contain

#30.0 color(red)(cancel(color(black)("cm"^3))) * overbrace("0.480 moles KOH"/(10^3 color(red)(cancel(color(black)("cm"^3)))))^(color(blue)("= 0.480 mol L"^(-1))) = "0.0144 moles KOH"#

#18.0 color(red)(cancel(color(black)("cm"^3))) * overbrace(("0.350 moles H"_2"SO"_4)/(10^3color(red)(cancel(color(black)("cm"^3)))))^(color(blue)("= 0.350 mol L"^(-1))) = "0.00630 moles H"_2"SO"_4#

Now, notice that you don't have enough moles of sulfuric acid to ensure that all the moles of potassium hydroxide will take part in the reaction because the sulfuric acid will only consume

#0.00630 color(red)(cancel(color(black)("moles H"_2"SO"_4))) * "2 moles KOH"/(1color(red)(cancel(color(black)("mole H"_2"SO"_4)))) = "0.0126 moles KOH"#

This means that after the reaction is complete, you will be left with #0# moles of sulfuric acid--in other words, sulfuric acid acts as a limiting reagent--and

#"0.0144 moles KOH " - " 0.0126 moles KOH" = "0.00180 moles KOH"#

The total volume of the solution will be

#"30.0 cm"^3 + "18.0 cm"^3 = "48.0 cm"^3#

The concentration of the hydroxide anions--keep in mind that potassium hydroxide is a strong base that dissociates in a #1:1# mole ratio* to produce hydroxide anions--in the resulting solution will be

#["OH"^(-)] = "0.00180 moles"/(48.0 * 10^(-3)color(white)(.)"L") = "0.0375 mol L"^(-)#

Note that I used the fact that #"1 cm"^3 = "1 mL" = 10^(-3)"L"# to convert the volume of the solution to liters!

Finally, the #"pH"# of an aqueous solution at room temperature is given by

#color(blue)(ul(color(black)("pH" = 14 + log(["OH"^(-)]))))#

Note that I got to this equation by using #"pOH" = - log(["OH"^(-)])# and #"pH + pOH" = 14#

Plug in your value to find

#"pH" = 14 + log(0.0375) = color(darkgreen)(ul(color(black)(12.574)))#

The answer is rounded to three decimal places, the number of sig figs you have for your values.