Volumetric and Gravimetric analysis with answers

Examples of Acids and Bases
Strong Acid
Hydrochloric HCl
Sulphuric H2SO4
Nitric HNO3
(Anything with a high KA value)
Weak Acid
Ammonium ion NH4
Benzoic C6H5COOH
Boric H3BO3
Ethanoic CH3COOH
Hydrocyanic HCN
(For more weak acids see page
11 of the data book)
Strong Base
Sodium Hydroxide NaOH
Potassium Hydroxide KOH
Barium Hydroxide Ba(OH)2
(most hydroxides)
Weak Base
Ammonia NH3
Alanine C3H5O2NH2
Methylamine CH3NH2
Pyridine C5H5N
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
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 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
V = 25.0 x 10-3L M(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 added
n(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 equation
therefore 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 sample
2HCl(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 sample
m = 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.
• 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 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.
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 500.0 mL volumetric flask.
the percentage, by mass, of magnesium sulfate in the powder.
Titration exam question
0.415 g of a pure acid, H2X(s), is added to exactly 100 mL of 0.105 M
A reaction occurs according to the equation
H2X(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.
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.
Back Titration exam question

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