VOLUMETRIC AND GRAVIMETRIC EXAM QUESTIONS

Examples of Acids and BasesStrong Acid Strong Base

Hydrochloric HClSulphuric H2SO4

Nitric HNO3(Anything with a high KA value)

Sodium Hydroxide NaOHPotassium Hydroxide KOHBarium Hydroxide Ba(OH)2

(most hydroxides)Weak Acid Weak Base

Ammonium ion NH4Benzoic C6H5COOH

Boric H3BO3Ethanoic CH3COOHHydrocyanic HCN

(For more weak acids see page 11 of the data book)

Ammonia NH3Alanine C3H5O2NH2

Methylamine CH3NH2Pyridine C5H5N

Titration example

30 mL of 0.10M NaOH neutralised 25.0mL of hydrochloric acid. Determine the concentration of the acidWrite 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 = ?

Titration example continued

Calculate moles NaOHn(NaOH) = M x V = 0.10 x 30 x 10-3 = 3 x 10-3 moles

From the balanced chemical equation find the mole ratioNaOH:HCl 1:1

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

Calculate concentration of HCl: M = n ÷ Vn = 3 x 10-

3 mol, V = 25.0 x 10-3LM(HCl) = 3 x 10-3 ÷ 25.0 x 10-3 = 0.12M

Back Titration example

A sample of calcium carbonate is dissolved with 20.00 mL of 0.2254 M hydrochloric acid and the excess acid is titrated with 0.1041 M sodium hydroxide. After dissolution, a mass of 0.2719 g of the calcium carbonate sample requires a titer of 12.39 mL of sodium hydroxide to reach a phenolphthalein end point. Find the %w/w of CaCO3 in the sample.

Back Titration continued• Find the total number of moles of HCl addedn(HCl total) = C(HCl) X V(HCl) = (0.2254 M)(0.020L) = 0.004508 mole HCl• Find the number of moles of NaOH needed to neutralise

the remaining HCl?• n(NaOH) = C(NaOH) X V(NaOH) = (0.1041 N)(0.01239 L) =

0.001290 mole NaOH• How much HCl is consumed by the NaOH titrant?1:1 ratio in balanced equationtherefore n(HCl in aliquot) =n(NaOH) =0.001290 mole HCl

Back Titration continued

• Find the number of moles of HCl consumed by the CaCO3

n(HCl consumed) = n(HCl total) – n(HCl aliquot)= 0.004508 – 0.001290 = 0.003218 mole

• Find the number of moles of CaCO3 in the sample2HCl(aq) + CaCO3(aq) CaCl2(aq) + H2O(aq) + CO2(g)

n(CaCO3) = ½ n(HCl) = ½ x 0.003218 = 0.001609 mole• Find the mass of CaCO3 in the samplem = n X M = 0.01609 X 100.1 = 0.1610g• Find the %w/w in the sample%w/w = m(CaCO3)/m(sample) X 100% =0.1610/0.2719 X 100% = 59.21% w/w CaCO3 in the sample

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• 32.50 g of the impure magnesium sulfate is dissolved in water and the

solution is made up to 500.0 mL in a volumetric fl ask.• Different volumes of 0.100 M BaCl2(aq) are added to six separate 20.00 mL

samples of this solution. This precipitates the sulfate ions as barium sulfate. The equation for the reaction isBa2+(aq) + SO4

2–(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.

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

i. the amount, in mole, of SO42–(aq) in the 500.0 mL volumetric flask.

ii. the percentage, by mass, of magnesium sulfate in the powder.

iii. .

iv.

Titration exam question0.415 g of a pure acid, H2X(s), is added to exactly 100 mL of 0.105 M NaOH(aq).A reaction occurs according to the equationH2X(s) + 2NaOH(aq) → Na2X(aq) + 2H2O(l)The NaOH is in excess. This excess NaOH requires 25.21 mL of 0.197 M HCl(aq) for neutralisation.

Calculatei. the amount, in mol, of NaOH that is added to the acid H2X initially.

ii.

ii. the amount, in mol, of NaOH that reacts with the acid H2X

iii. the molar mass, in g mol–1, of the acid H2X.

Back Titration exam question