Need help finding pH?

Question

What is the pH at the point during a titration when 29 mL of 0.200 M HCl has been added to 27 mL of 0.0500 M NH3 solution? HCl is strong acid, Kb= 1.8 x 10^ -5 for NH3.

Answer 1

In the titration you can assume the since HCl is a strong acid it reacts completly with NH3. You use stoichiometry to determine the amount of acid or base left after the addition of titrant. To do this you need to look at the chemical equation.

HCl(aq) + NH3(aq) --> NH4+(aq) + Cl-(aq)

For every one mole of HCl that reacts one mole of NH3 is used.

.029L of HCl *(.2moles/L) = .0058moles of HCl

.027L of NH3 *(.05moles/L) = .00135moles of NH3

Because HCl is in excess, the NH3 will be the limiting reagent in the reaction. Now use mole ratios to determine the amount of HCl that react with the base.

.00135moles of NH3 *(1mole HCl/1mole NH3)= .00135moles of HCl that reacted

To calculate the amount of HCl left after the reaction, simply subtract the moles of HCl total from the amount of HCl that reacted.

(.0058moles of HCl total) - (.00135moles of HCl reacted) = .00445moles of HCl left

Now calculate the concentration of HCl in the solution.

.00445moles of HCl *(mole/.056L) = .07946M of HCl

HCl is a strong acid so it ionizes completly

HCl(aq) --> H+(aq) + Cl-(aq)

So for every single HCl molecule that ionizes, one hydrogen ion is produced.

[HCl] = [H+] = .07946M  

pH= -log[H+]

pH= -log(.07946)

pH= 1.10

Answer 2

HCl + NH3 = NH4+ + Cl-

mol HCl = 0.029*0.2 = 0.0058 mol

mol NH3 = 0.027 * 0.05 = 0.00135 mol

note that in the stoichiometric eqn. 1 mol HCl completely reacts with 1 mol NH3

Therefore HCl is in excess and NH3 is the limiting reagent.

excess HCl = 0.0058 - 0.00135 = 0.00445 mol

After reaction 0.00445 mol HCl remains in a solution of total volume (29+27)mL

1 mol HCl forms 1 mol H+

[H+] = 0.00445 / 0.056 = 0.0795 M

Hence pH = - log (0.0795) = 1.1