Question #ddd5c

You're mixing a solution of hydrochloric acid, #"HCl"#, a strong acid, with a solution of potassium hydroxide, #"KOH"#, a strong base, so right from the start you know that the a complete neutralization would produce a neutral solution, i.e. a solution of #"pH" = 7#.

Your starting point here will be balanced chemical equation that describes this neutralization reaction

#"HCl"_ ((aq)) + "KOH"_ ((aq)) -> "KCl"_ ((aq)) + "H"_ 2"O"_ ((l))#

If you eliminate the spectator ions, you can write the net ionic equation

#"H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l))#

Here every mole of hydronium cations, #"H"_3"O"^(+)#, is produced by #1# mole of hydrochloric acid and every mole of hydroxide anions, #"OH"^(-)#, is produced by #1# mole of potassium hydroxide.

This tells you that in order to have a complete neutralization, you need to have equal numbers of moles of hydronium cations and hydroxide anions.

Use the molarities and volumes of the two solutions to calculate how many moles of each you're mixing

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#

For the potassium hydroxide solution you will have

#n_("OH"^(-)) = "0.183 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(45.0 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))#

#= "0.008235 moles OH"^(-)#

For the hydrochloric acid solution you will have

#n_("H"_3"O"^(+)) = "0.145 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(65.0 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))#

#= "0.009425 moles H"_3"O"^(+)#

Notice that you have more moles of hydronium cations than of hydroxide anions. The reaction consumes the two in a #1:1# mole ratio, which means that the moles of hydroxide anions will be completely consumed by the reaction.

The resulting solution will contain

#n_("OH"^(-)) = "0 moles " -># completely consumed

#n_("H"_3"O"^(+)) = "0.009425 moles" - "0.008235 moles"#

#= "0.001190 moles H"_3"O"^(+)#

Now, the pH of the solution is defined as

#color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))#

Here

#["H"_3"O"^(+)]# - the concentration of hydronium cations

In order to determine the concentration of hydronium cations, calculate the total volume of the resulting solution

#V_ "total" = V_("H"_ 3"O"^(+)) + V_("OH"^(-))#

#V_"total" = "65.0 mL" + "45.0 mL" = "110.0 mL"#

The concentration of hydronium cations will thus be

#["H"_3"O"^(+)] = "0.001190 moles"/(110.0 * 10^(-3)"L") = "0.01082 mol L"^(-1)#

The pH of the solution will be

#"pH" = - log(0.01082) = color(green)(|bar(ul(color(white)(a/a)color(black)(1.97)color(white)(a/a)|)))#

So, does this result make sense?

Since you don't have enough strong base to completely neutralize the strong acid, you should expect the resulting solution to be quite acidic, i.e. #"pH" < 7#.