Titration of Weak Acid with Strong Base?

Question

A certain weak acid, HA, with a Ka of 5.61 x 10^-6, is being titrated with a strong base. A solution is made by mixing 9.00 millimoles of HA and 3.00 millimoles of strong base. What is the resulting pH? Thanks in advance!

Answer 1

You have to indicate the volume of the two solutions.
In any case if we suppose 10 mL of acid mixed wiyh 10 mL of base this is the solution.

9 millimoles =0.0009 mole HA

3 millimoles = 0.0003 mole OH-

10 mL= 0.01 L

Total volume = 20 mL=0.02 L

The acid and base react according to the net equation

HA + OH- >> H2O + A-

From 0.003 mole OH- plus 0.009 mole Ha we would get enought reaction to use up 0.003 mole HA leaving 0.006 mole HA and forming 0.003 mole A-
Concentration HA = 0.006 / 0.02 = 0.3 M
Concentration A- = 0.003 / 0.02 = 0.15 M
At equilibrium conc. HA = 0.3-x Conc H+ = x
Conc A-=0.15+x
5.61 10^-6 = (x)(0.15+x) / 0.3-x
x= 0.0000112 = conc. H+
pH = -log conc H+ = 4.95

Answer 2

maurizia is right from the point of view that the exact solution, as provided in her answer, requires that you know the volumes of the solutions.

If you don't know them, then you can only get an approximate solution using the Henderson -Hasselbach equation.

pH= pKa + log( [conjugate base] / [acid]) =>
pH= pKa +log( [A-]/[HA] )
If the final volume is V then [A-] =mole A-/V
and [HA] =mole HA/V Thus in the ratio the volume is simplified. Also the conversion factor from mole to mmole is simplified thus you have

pH= pKa+log (mmole A- / mmole HA)

I assume we are talking about a monovalent base since usually titrants are monovalent bases but also it is the simplest case.

The reaction is 1:1 so you will have
forming mmole A- = mmole strong base =3 mmole and
remaining mmole HA = mmole HA (initial) -mmole strong base = 9-3 =6

So pH=-log(5.61*10^-6) + log(3/6) =4.95

You should note that this equation is exactly the same as maurizia wrote, only formulated a bit differently. Also I am doing here the approximation that the concentations of HA and A- at equilibrium are practically the same as the initial concentrations at the moment the two solutions mixed. This approximation is frequently done and judging from the formulation of your question this is what your teacher expects from you.