Question #0c050

As you know, the #"pH"# of the solution is given by the concentration of hydronium cations, #"H"_ 3"O"^(+)#, present in the solution.

Now, hydrochloric acid and sodium hydroxide neutralize each other in a #1:1# mole ratio to produce water and aqueous sodium chloride.

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

Your starting point here will be to figure out if you're dealing with a complete neutralization, in which case, the #'pH"# of the solution will be equal to that of pure water at room temperature, i.e. to #7#.

Use the molarities and volume of the two solutions to calculate how many moles are mixed. You will have

#9.9 color(red)(cancel(color(black)("mL solution"))) * "1 mole HCl"/(10^3color(red)(cancel(color(black)("mL solution")))) = 9.9 * 10^(-3)color(white)(.)"moles HCl"#

for the hydrochloric acid solution and

#100 color(red)(cancel(color(black)("mL solution"))) * "0.1 moles NaOH"/(10^3color(red)(cancel(color(black)("mL solution")))) = 10 * 10^(-3)color(white)(.)"moles NaOH"#

for the sodium hydroxide solution. As you can see, you have more moles of sodium hydroxide than of hydrochloric acid, which implies that sodium hydroxide is in excess, or, in other words, that hydrochloric acid is the limiting reagent.

You can thus say that the reaction will consume #9.9 * 10^3# moles of hydrochloric acid and of sodium hydroxide and leave behind

#10 * 10^(-3)color(white)(.)"moles" - 9.9 * 10^(-3)color(white)(.)"moles" = 0.1 * 10^(-3)color(white)(.)"moles"#

of sodium hydroxide. Since sodium hydroxide is a strong base, you can say that the resulting solution will contain #0.1 * 10^(-3)# moles of hydroxide anions, #"OH"^(-)#.

The total volume of the solution will be

#"9.9 mL + 100 mL = 109.9 mL"#

Now, the concentration of hydroxide anions

#["OH"^(-)] = (0.1 * 10^(-3)color(white)(.)"moles")/(109.9 * 10^(-3)color(white)(.)"L")#

will give you the #"pOH"# of the solution.

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

As you know, an aqueous solution at room temperature has

#"pH" = 14 - "pOH"#

This means that the #"pH"# of the resulting solution will be

#"pH" = 14 - [- log((0.1 * 10^(-3)color(white)(.)"moles")/(109.9 * 10^(-3)color(white)(.)"L"))]#

#color(darkgreen)(ul(color(black)("pH" = 11.0)))#

The answer is rounded to one decimal place, the number of sig figs you have for most of your values.