Question #753db

I'll focus on parts (c) and (d), since they're more interesting, and leave parts (a) and (b) to you as practice.

So, for part (c), you know that you're dealing with a solution of potassium hydroxide, a strong base, which you dilute from an initial volume of 10 mL to a final volume of 250 mL.

Use the molarity and initial volume of the solution to find how many moles of potassium hydroxide you're working with

#color(blue)(C = n/V implies n = C * V)#

#n = "0.22 M" * 10.0 * 10^(-3)"L" = "0.0022 moles"#

Since potassium hydroxide is a strong base, it will dissociate completely in aqueous solution to form

#"KOH"_text((aq]) -> "K"_text((aq])^(+) + "OH"_text((aq])^(-)#

Notice that every mole of potassium hydroxide will produce one mole of hydroxide ions. This means that you have

#n_(OH^(-)) = n_(KOH) = "0.0022 moles"#

The molarity of the hydroxide ions after the solution is diluted is

#["OH"^(-)] = "0.0022 moles"/(250 * 10^(-3)"L") = "0.0088 M"#

This means that the #pOH# of the solution will be

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

#pOH = -log(0.0088) = color(green)(2.06)#

The pH of the solution will be

#pH_"sol" = 14 - pOH = 14 - 2.06 = color(green)(11.94)#

For part (d), you have to use the molar mass of barium hydroxide, #"Ba"("OH")_2#, to find the number of moles you have in that mass.

#0.50color(red)(cancel(color(black)("g"))) * "1 mole"/(171.34color(red)(cancel(color(black)("g")))) = "0.00292 moles"#

Barium hydroxide will dissociate in aqueous solution to form

#"Ba"("OH")_text(2(aq]) -> "Ba"_text((aq])^(2+) + color(red)(2)"OH"_text((aq])^(-)#

This time, every mole of barium hydroxide will produce #color(red)(2)# moles of hydroxide ions, which means that you have

#n_(OH^(-)) = color(red)(2) * n_(Ba(OH)_2) = 2 * 0.00292 = "0.00584 moles"#

The molarity of the hydroxide ions will thus be

#["OH"^(-)] = "0.00584 moles"/(100 * 10^(-3)"L") = "0.0584 M"#

The solution's #pOH# is

#pOH = -log(0.0584) = color(green)(1.23)#

This means that the pH is

#pH_"sol" = 14 - 1.23 = color(green)(12.77)#