Calculate the pH during a titration when 9.54 mL of a 0.15 M HCl solution has reacted with 22.88 mL of 0.14 M NaOH?

Start by calculating the number of moles of hydrochloric acid and the number of moles of sodium hydroxide that took part in the reaction.

#9.54 color(red)(cancel(color(black)("mL solution"))) * "0.15 moles HCl"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.001431 moles HCl"#

#22.88 color(red)(cancel(color(black)("mL solution"))) * "0.14 moles NaOH"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.003203 moles NaOH"#

Now, hydrochloric acid and sodium hydroxide neutralize each other in a #1:1# mole ratio.

#"HCl"_ ((aq)) + "NaOH"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O"_ ((l))#

This tells that in order to have a complete neutralization, you need to mix equal numbers of moles of hydrochloric acid and of sodium hydroxide.

In your case, you have more moles of sodium hydroxide than of hydrochloric acid, so you can say that the acid will act as a limiting reagent, i.e. it will be completely consumed before all the moles of sodium hydroxide will get the chance to react.

This means that after the reaction takes places, your solution will contain #0# moles of hydrochloric acid and

#"0.003203 moles " - " 0.001431 moles" = "0.001772 moles NaOH"#

At this point, you should be able to say that the #"pH"# of the resulting solution must be #>7# because you're left with a solution that contains a strong base.

As you know, sodium hydroxide dissociates completely in aqueous solution to produce sodium cations and hydroxide anions.

#"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)#

This means that the resulting solution will contain #0.0017772# moles of hydroxide anions.

The total volume of the resulting solution will be

#"9.54 mL + 22.88 mL = 32.42 mL"#

The molarity of the hydroxide anions will be--don't forget to convert the volume of the solution to liters!

#["OH"^(-)] = "0.001772 moles"/(32.42 * 10^(-3) quad "L") = "0.05466 mol L"^(-1)#

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

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

Since you know that

#color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)])))#

you can say that

#"pH" = 14 + log(["OH"^(-)])#

Plug in your value to find

#"pH" = 14 + log(0.05466) = color(darkgreen)(ul(color(black)(12.74)))#

The answer is rounded to two decimal places, the number of sig figs you have for the concentrations of the two solutions.