CH 117 Spring 2015Worksheet 14

  1. What is the difference between the equivalence point and the end point of a titration?

The equivalence point is a theoretical, calculated value where there are exactly equal amounts of acid and base present in the titration solution. The end point is the experimental point at which the amount of acid and base should be the same, detected in the lab by indicators. If you have done the titration correctly and selected an appropriate indicator, the equivalence point and end point will be the same.

  1. Draw and label the titration curve for the titration of a strong acid with a strong base.

  1. Draw and label the titration curve for the titration of a strong base with a strong acid.

  1. What is the pH at the point in a titration at which 20.00 mL of 1.0 M KOH has been added to 25.00 mL of 1.0 M HBr?

First identify the type of titration – this is a strong acid titrated with a strong base, so the first curve we drew.

Strong acid/strong base titrations are nothing more than limiting reagent problems, so we’ll use a limiting reagent ICE table to visualize what’s happening. We really only care about the reactants side for strong acid/strong base titrations since the salt and the water that are produced will both HAVE NO EFFECT on pH. Titration limiting reagent ICE charts must be worked in units of moles, so make sure to multiply volume by molarity for each chemical to get started.

Moles of HBr = .025 L x 1.0 M = .025 moles HBr

Moles of KOH = .020 L x 1.0 M = .02 moles KOH

Since there is less of the KOH to begin with, that will be our limiting reagent. Remember to use the ICE chart to get rid of all of the limiting reagent and subtract/add from the other reagents as appropriate.

HBr / KOH /  / KBr / H2O
Initial / .025 moles / .02 moles / n/a / n/a
Change / -.02 moles / -.02 moles / n/a / n/a
Equilibrium (Final) / .005 moles / 0 moles / n/a / n/a

So, our final solution has .005 moles of HBr remaining. Since HBr is a strong acid, we can calculate pH simply by taking the –log of the concentration of acid left.

Since we have moles, we must divide by the FINAL TOTAL VOLUME to calculate molarity of acid remaining. This is an important step in titration problems. Do not forget that you are working in moles, so it is necessary to calculate a concentration before finding the pH or pOH.

[HBr] = [H3O+] = .005 moles / (.025 + .02 L) = .111 M

pH = -log[H3O+] = -log(.111) = 0.95

  1. What is the pH at the point in a titration at which 55.00 mL of 1.0 M KOH has been added to 25.00 mL of 1.0 M HBr?

Identify type of titration – strong acid + strong base

The process involved is identical to the process in #4 with different numbers.

Moles of HBr = .025 L x 1.0 M = .025 moles HBr

Moles of KOH = .055 L x 1.0 M = .055 moles of KOH

HBr is our limiting reagent now.

HBr / KOH /  / KBr / H2O
Initial / .025 moles / .055 moles / n/a / n/a
Change / -.025 moles / -.025 moles / n/a / n/a
Equilibrium / 0 moles / .03 moles / n/a / n/a

Final solution has .03 moles of KOH  [KOH] = .03 moles/ (.025 L + .055 L) = .375 M

[KOH] = [OH-} = .375 M

pOH = -log(.375) = .43

pH = 14 - .43 = 13.6