Chemistry question over acid and bases? I am not sure what I am doing wrong.

My first guess would be that your answer was marked as incorrect because you used the wrong number of significant figures--three instead of two--but I'll go through the whole calculation again just for good measure.

The idea here is that you need to use the number of moles of hydroxide anions present in the two solutions to figure out the concentration of the hydroxide anions in the resulting solution.

You know that

#color(blue)(ul(color(black)(["OH"^(-)] = 10^(-"pOH") quad "M")))#

so you can say that the first solution will have

#["OH"^(-)]_ 1 = 10^(-3.00) quad "M"#

#["OH"^(-)]_ 1 = 1.00 * 10^(-3) quad "M"#

Similarly, the second solution will have

#["OH"^(-)]_ 2 = 10^(-3.50)quad "M"#

#["OH"^(-)]_ 2 = 3.16 * 10^(-4)quad "M"#

The number of moles of hydroxide anions present in the first solution is equal to

#2.0 * 10^2color(red)(cancel(color(black)("mL solution"))) * (1.00 * 10^(-3) quad "moles OH"^(-))/(10^3color(red)(cancel(color(black)("mL solution")))) = 2.00 * 10^(-4) quad "moles OH"^(-)#

Similarly, the number of moles of hydroxide anions present in the second solution is equal to

#1.600 color(red)(cancel(color(black)("L solution"))) * (3.16 * 10^(-4) quad "moles OH"^(-))/(1color(red)(cancel(color(black)("L solution")))) = 5.06 * 10^(-4) quad "moles OH"^(-)#

This means that the total number of moles of hydroxide anions present in the resulting solution will be

#(2.00 * 10^(-4) + 5.06 * 10^(-4))quad "moles OH"^(-) = 7.06 * 10^(-4) quad "moles OH"^(-)#

The total volume of the resulting solution will be--do not forget to mix liters with liters or milliliters with milliliters!

#(2.0 * 10^(2) * 10^(-3) + 1.6) quad "L" = "1.8 L"#

This means that the concentration of hydroxide anions in the resulting solution will be

#["OH"^(-)]_ "final" = (7.06 * 10^(-4) quad "moles")/("1.8 L") = color(darkgreen)(ul(color(black)(3.9 * 10^(-4) quad "M")))#

The answer must be rounded to two sig figs, the number of sig figs you have for the volume of the first solution.

Now, you know that an aqueous solution at #25^@"C"# has

#color(blue)(ul(color(black)("pH + pOH" = 14)))#

and

#color(blue)(ul(color(black)(["H"_3"O"^(+)] * ["OH"^(-)] = 1 * 10^(-14) quad "M"^2)))#

This means that the resulting solution will have

#["H"_3"O"^(+)] = (1 * 10^(-14) "M"^color(red)(cancel(color(black)(2))))/(3.9 * 10^(-4)color(red)(cancel(color(black)("M")))) = color(darkgreen)(ul(color(black)(2.6 * 10^(-11) quad "M")))#

Once again, the answer is rounded to two sig figs.

Now, you can say that the resulting solution will have

#"pOH" = - log(["OH"^(-)]_ "final")#

#"pOH" = - log(3.9 * 10^(-4)) = color(darkgreen)(ul(color(black)(3.41)))#

This time, the result must be rounded to two decimal places because you have two significant figures for the concentration of hydroxide anions.

Finally, you get

#"pH" = 14 - 3.41 = color(darkgreen)(ul(color(black)(10.59)))#

Once again, round the answer to two decimal places.