go to the AUS-e-TUTE homepage
home test Join AUS-e-TUTE game contact
 

 

Acid-base Titration Calculations

Key Concepts

  • acid-base reactions involve a proton transfer

  • the acid donates a proton to the base

  • acid-base reactions are also known as neutralisation reactions

    acid + base → salt + water

    acid A + base B → conjugate acid of base B + conjugate base of acid A
    (Lowry-Brönsted theory)

  • H+ + OH- → H2O is the most general neutralisation reaction

  • Equivalence point is the point at which the moles of H+ is equal to the moles of OH-
        An indicator is used to show the equivalence point during a titration

  • A titration involves the progressive addition of one reactant from a burette
        (usually the acid), to a known volume of the other reactant in a conical flask
        (usually the base)

Calculations

  1. Write the balanced chemical equation for the reaction

  2. Extract all the relevant information from the question

  3. Check that data for consistency, for example, concentrations are usually given in M or mol L-1 but volumes are often given in mL. You will need to convert the mL to L for consistency. You need to divide the volume in mL by 1000, and the easiest way to do this on your calculator is to multiply the volume in mL x 10-3 (enter the volume in mL into your calculator, then click "EXP" then -3, and you've done the coversion to L !)

  4. Calculate the moles of reactant (n) for which you have both the volume(V) and concentration(c) : n = c x V

  5. From the balanced chemical equation find the mole ratio known reactant : unknown reactant

  6. Use the mole ratio to calculate the moles of the unknown reactant

  7. From the volume(V) of unknown reactant and its previously calculated moles(n), calculate its concentration(c): c = n ÷ V

Examples

1. 30 mL of 0.10 mol L-1 NaOH neutralised 25.0 mL of hydrochloric acid.
    Determine the concentration of the acid

  1. Write the balanced chemical equation for the reaction
          NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

  2. Extract the relevant information from the question:
          NaOH      V = 30 mL , M = 0.10 mol L-1
          HCl      V = 25.0 mL, c = mol L-1 ?

  3. Check the data for consistency
          NaOH      V = 30 x 10-3 L , c = 0.10 mol L-1
          HCl      V = 25.0 x 10-3 L, c = mol L-1 ?

  4. Calculate moles NaOH
          n(NaOH) = c x V = 0.10 x 30 x 10-3 = 3 x 10-3 moles

  5. From the balanced chemical equation find the mole ratio
          NaOH:HCl
          1:1

  6. Find moles HCl
          NaOH: HCl is 1:1
          So n(NaOH) = n(HCl) = 3 x 10-3 moles at the equivalence point

  7. Calculate concentration of HCl: c = n ÷ V
          n = 3 x 10-3 mol,       V = 25.0 x 10-3 L
          c(HCl) = 3 x 10-3 ÷ 25.0 x 10-3 = 0.12 M or 0.12 mol L-1

2. 50mL of 0.2 mol L-1 NaOH neutralised 20 mL of sulfuric acid.
    Determine the concentration of the acid.

  1. Write the balanced chemical equation for the reaction
          2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)

  2. Extract the relevant information from the question:
          NaOH       V = 50 mL, c = 0.2 mol L-1
          H2SO4       V = 20 mL, c = mol L-1?

  3. Check the data for consistency
          NaOH       V = 50 x 10-3 L, c = 0.2 mol L-1
          H2SO4       V = 20 x 10-3 L, c = mol L-1?

  4. Calculate moles NaOH
          n(NaOH) = c x V = 0.2 x 50 x 10-3 = 0.01 mol

  5. From the balanced chemical equation find the mole ratio
          NaOH:H2SO4
          2:1

  6. Find moles H2SO4
          NaOH: H2SO4 is 2:1
          So n(H2SO4) = ½ x n(NaOH) = ½ x 0.01 = 5 x 10-3 moles H2SO4 at the equivalence point

  7. Calculate concentration of H2SO4: c = n ÷ V
          n = 5 x 10-3 mol,       V = 20 x 10-3L
          c(H2SO4) = 5 x 10-3 ÷ 20 x 10-3 = 0.25 M or 0.25 mol L-1

3. 25.0 mL of 0.05 mol L-1 Ba(OH)2 neutralised 40.0 mL of nitric acid.
    Determine the concentration of the acid.

  1. Write the balanced chemical equation for the reaction
          Ba(OH)2(aq) + 2HNO3(aq) → Ba(NO3)2(aq) + 2H2O(l)

  2. Extract the relevant information from the question:
          Ba(OH)2       V = 25.0 mL, M = 0.05 mol L-1
          HNO3       V = 40.0 mL, c = mol L-1?

  3. Check the data for consistency
          Ba(OH)2       V = 25.0 x 10-3 L, c = 0.05 mol L-1
          HNO3       V = 40.0 x 10-3 L, c = mol L-1?

  4. Calculate moles Ba(OH)2
          n = c x V = 0.05 x 25.0 x 10-3 = 1.25 x 10-3 mol

  5. From the balanced chemical equation find the mole ratio
          Ba(OH)2 : HNO3
          1 : 2

  6. Find moles HNO3
          Ba(OH)2 : HNO3 is 1 : 2
          So n(HNO3) = 2 x n(Ba(OH)2) = 2 x 1.25 x 10-3 = 2.5 x 10-3 mol

  7. Calculate concentration of HNO3: c = n ÷ V
          n = 2.5 x 10-3moles, V = 40.0 x 10-3 L
          c = 2.5 x 10-3 ÷ 40.0 x 10-3 = 0.0625 M or 0.0625 mol L-1


What would you like to do now?
advertise on the AUS-e-TUTE website and newsletters
 
 

Search this Site

You can search this site using a key term or a concept to find tutorials, tests, exams and learning activities (games).
 

Become an AUS-e-TUTE Member

 

AUS-e-TUTE's Blog

 

Subscribe to our Free Newsletter

Email email us to
subscribe to AUS-e-TUTE's free quarterly newsletter, AUS-e-NEWS.

AUS-e-NEWS quarterly newsletter

AUS-e-NEWS is emailed out in
December, March, June, and September.

 

Ask Chris, the Chemist, a Question

The quickest way to find the definition of a term is to ask Chris, the AUS-e-TUTE Chemist.

Chris can also send you to the relevant
AUS-e-TUTE tutorial topic page.

 
 
 

Share this Page

Bookmark and Share
 
 

© AUS-e-TUTE