1. VOLUMETRIC AND GRAVIMETRIC EXAM QUESTIONS

2. Examples of Acids and Bases
Strong Acid
Strong Base
Hydrochloric HCl
Sulphuric H2SO4
Nitric HNO3
(Anything with a high KA value)
Sodium Hydroxide NaOH
Potassium Hydroxide KOH
Barium Hydroxide Ba(OH)2
(most hydroxides)
Weak Acid
Weak Base
Ammonium ion NH4
Benzoic C6H5COOH
Boric H3BO3
Ethanoic CH3COOH
Hydrocyanic HCN
(For more weak acids see page 11 of the data book)
Ammonia NH3
Alanine C3H5O2NH2
Methylamine CH3NH2
Pyridine C5H5N

3. Titration example
30 mL of 0.10M NaOH neutralised 25.0mL of hydrochloric acid. Determine the concentration of the acid Write the balanced chemical equation for the reaction NaOH(aq) + HCl(aq) -----> NaCl(aq) + H2O(l) Extract the relevant information from the question: NaOH V = 30mL , M = 0.10M HCl V = 25.0mL, M = ? Check the data for consistency NaOH V = 30 x 10-3L , M = 0.10M HCl V = 25.0 x 10-3L, M = ?

4. Titration example continued
Calculate moles NaOH n(NaOH) = M x V = 0.10 x 30 x 10-3 = 3 x 10-3 moles From the balanced chemical equation find the mole ratio NaOH:HCl 1:1 Find moles HCl NaOH: HCl is 1:1 So n(NaOH) = n(HCl) = 3 x 10-3 moles at the equivalence point Calculate concentration of HCl: M = n ÷ V n = 3 x 10-3 mol, V = 25.0 x 10-3L M(HCl) = 3 x 10-3 ÷ 25.0 x 10-3 = 0.12M

5. Back Titration example
A sample of calcium carbonate is dissolved with mL of M hydrochloric acid and the excess acid is titrated with M sodium hydroxide. After dissolution, a mass of g of the calcium carbonate sample requires a titer of mL of sodium hydroxide to reach a phenolphthalein end point. Find the %w/w of CaCO3 in the sample.

6. Back Titration continued
Find the total number of moles of HCl added
n(HCl total) = C(HCl) X V(HCl) = ( M)(0.020L)
= mole HCl
Find the number of moles of NaOH needed to neutralise the remaining HCl?
n(NaOH) = C(NaOH) X V(NaOH) = ( N)( L) = mole NaOH
How much HCl is consumed by the NaOH titrant?
1:1 ratio in balanced equation
therefore n(HCl in aliquot) =n(NaOH) = mole HCl

7. Back Titration continued
Find the number of moles of HCl consumed by the CaCO3
n(HCl consumed) = n(HCl total) – n(HCl aliquot)
= – = mole
Find the number of moles of CaCO3 in the sample
2HCl(aq) + CaCO3(aq)  CaCl2(aq) + H2O(aq) + CO2(g)
n(CaCO3) = ½ n(HCl) = ½ x = mole
Find the mass of CaCO3 in the sample
m = n X M = X = g
Find the %w/w in the sample
%w/w = m(CaCO3)/m(sample) X 100%
=0.1610/ X 100% = 59.21% w/w CaCO3 in the sample

8. Exam questions on Gravimetric and volumetric analysis
The percentage purity of powdered, impure magnesium sulfate, MgSO4, can be determined by gravimetric analysis. Shown below is the method used in one such analysis. Method • g of the impure magnesium sulfate is dissolved in water and the solution is made up to mL in a volumetric fl ask. • Different volumes of M BaCl2(aq) are added to six separate mL samples of this solution. This precipitates the sulfate ions as barium sulfate. The equation for the reaction is Ba2+(aq) + SO42–(aq) → BaSO4(s) The precipitate from each sample is filtered, rinsed with de-ionised water and then dried to constant mass. The results of this analysis are shown on the next page.

10. a. Why is it necessary to rinse the precipitate with de-ionised water before drying?
b. Explain why the amount of BaSO4(s) precipitated remains constant for the last four samples tested even though more BaCl2(aq) is being added.
c. Calculate
the amount, in mole, of SO42–(aq) in the mL volumetric flask.
the percentage, by mass, of magnesium sulfate in the powder.

12. Titration exam question
0.415 g of a pure acid, H2X(s), is added to exactly 100 mL of M NaOH(aq).
A reaction occurs according to the equation
H2X(s) + 2NaOH(aq) → Na2X(aq) + 2H2O(l)
The NaOH is in excess. This excess NaOH requires mL of M HCl(aq) for neutralisation.
Calculate
the amount, in mol, of NaOH that is added to the acid H2X initially.
the amount, in mol, of NaOH that reacts with the acid H2X
the molar mass, in g mol–1, of the acid H2X.

13. Back Titration exam question