Back titrations

Could someone help me with this question?

In order to find the concentration of some dilute HCl acid, a student took 25cm3 and diluted it accurately to 250cm3. She made up a standard solution of sodium hydrogencarbonate by dissolving 2.25g of it in water, and making up the solution to 250cm3. A 25cm3 sample of this solution required 25.6cm3 of the diluted HCl acid to react to completion. Find the concentration of the original acid.

Eqn: NaHCO3 + HCl --> NaCl + CO2 + H2O

I tried this and got some really funny answers. Any help is appreciated! :smile:

Reply 1

Dystopia

Let C_a1 be the concentration of the original acid. Therefore:

C_a = C_a1 / 10

Concentration of NaHCO₃ = n/V = m/VN = 2.25/(0.25 x 84) = 0.107 mol dm^-3
Note that the concentration of the sample will be the same.

C_a1/10 x 0.0256dm³ = 0.025dm³ x 0.107 mol dm^-3

C_a1 =

\frac{10 \; \times \; 0.025dm^3 \; \times \; 0.107 mol \; dm^{-3}}{0.0256dm^{3}}

So the value I get is 1.04 mol dm^-3.

I haven't checked all that through though, and titrations are just annoying. :p:

Reply 2

Excalibur

Aha, thanks! :smile:

Gosh I'm stuck on another question... it sounds really complicated:

The mineral cerussite consists of lead carbonate, PbCO3. A particular lead ore consisted only of cerussite and inert rocky material. 1g of crushed ore was added to 25cm3 of 1 mol/dm3 nitric acid (an excess) in a beaker. When no more reaction was observed, the mixture was filtered and washing from the beaker were also run through the filter paper. The residue on the filter paper was washed thoroughly and the washings added to the filtrate. The solution formed was made up to 250cm3 with water. A 25cm3 sample of this was titrated with 0.1mol/dm3 sodium hydroxide. 20.5cm3 was required to produce a neutral solution. Calculate the % purity of lead in the ore.

All the filtering stuff really confused me. Could anyone give me a hand?

Reply 3

Eau

There are two reactions here:

1. PbCO3(s) + 2HNO3(aq) ---> Pb(NO3)2(aq) + CO2(g) + H2O(l)

This occurs when the ore is added to the nitric acid. The inert material does not react. Note that Pb(NO3)2 is soluble - it is contained in the filtrate.

2. 2NaOH(aq) + Pb(NO3)2(aq) ---> 2Na(NO3)(aq) + Pb(OH)2(s)

This is the neutralisation reaction when 25 cm^3 of Pb(NO3)2 is neutralised by 20.5 cm^3 of 0.1 M NaOH.

We have to work backwards (from now on a calculator and a periodic table will be handy):

Amount of NaOH used = 0.1 mol.dm^-3 * (20.5/1000) dm^3 = 0.00205 mol

Since NaOH and Pb(NO3)2 react in a 2:1 ratio,

Amount of Pb(NO3)2 in 25 cm^3 = 0.00205/2 = 0.001025 mol

Amount of Pb(NO3)2 in 250 cm^3 = 0.001025 * 10 = 0.01025 mol

Since 1 mole of Pb(NO3)2 can be produced from 1 mole of PbCO3,

Amount of PbCO3 in 1 g of rock = 0.01025 mol

Mass of PbCO3 = 0.01025 mol * (207 + 12 + 16*3) g.mol^-1 = 2.74 g

Hmmm.... are you sure it's 1 g of crushed ore, and not a bit more, say 3 g?

Ok... I might've made a mistake. I will check later - gtg!

Reply 4

Theorist

Excalibur, what board is this?

Reply 5

Excalibur

We're doing Edexcel. Why do you ask? :smile:

Reply 6

Theorist

Excalibur

We're doing Edexcel. Why do you ask? :smile:

Because we're doing Edexcel, but we certainly haven't been given any questions like this in our school - and we've done back titrations in class. :frown:

Were these questions from a textbook or something, because I'd quite like to get some practice (if that's the sort of thing we're going to be asked in the exam)? Would this come under Unit 3B or Unit 1 - Topic 1.2: Formulae, Equations and Moles?