Back Titrations

I need help with these:

1. 1.41g of a group 2 carbonate, MCOˇ3, were dissolved in 50cm³ of 1.0M HCL acid. After the reaction was complete the solution was made up to 250cm³ with distilled water in a volumetric flask. A 25cm³ portion of the solution was titrated with 01.M NaOH of which 31.0cm³ were required to neutralise the excess acid. Use these figures to calculate the relative formula mass of the carbonate MCOˇ3, and thence the identity of the metal M.

2. 0.467g of a divalent metal M was dissolved in 5cm³ of 2M sulphuric acid. Write the equation for the reaction of metal M with dilute sulphuric acid. The solution was then made up to 25cm³. The excess acid in this solution was titrated against 0.2M potassium hydroxide of which 27cm³ was required. Calculate the relative atomic mass of the metal.

3. Metal Z has a RAM of 24. 0.57g of Z was allowed to react with 50cm³ of 0.5M HCL acid. The excess of acid required 29.9cm³ of 0.096M potassium hydroxide solution for neutralisation. Use these data to calculate the number of moles of Z reacting with one mole of sulphuric acid and hence deduce the charge on the ion.

4. 1.60g of a metallic oxide of formula MO was dissolved in 100cm³ of 1.0M HCL acid. The resulting solution was made up to 500cm³ with distilled water. 25cm³ of this dilute solution required 21.03cm³ of 0.102M sodium hydroxide for neutralisation. Calculate the mass of the oxide reacting with one mole of HCL acid and hence the RAM of the metal

Any help would be greatly appreciated!

Reply 1

marcusfox

I'll give 1 a bash, besides, it should give you some idea on how to do the others. I'm assuming you mean 0.1M NaOH when you say 01.M NaOH.

Basically you need to determine how much of your metal carbonate in 1) is metal, and how much is carbonate. The ratio at which they will be combined for this is 1:1 metal:carbonate (M²⁺, CO₃^2-) You can do this by finding out how much carbonate is in the metal carbonate that is used in the titration, as this will react to form H₂CO₃, which is a ratio of 2:1

50 ml 1M HCl made up to 250 ml is a 5x dilution

HCl reacts with NaOH in the ratio 1:1. Find the number of moles in the 31 ml of NaOH that you titrated. Use mol dm^-3 as they're standard units. 1 dm³ is basically 1000 ml, or one litre.

0.1M x 31/1000 = 0.0031 moles. Therefore, the same number of moles of HCl reacted. So in that 25 ml of HCl there were 0.0031 moles.

Moles of HCl in 50 ml of a 1M standard solution = 1 x (50/1000) or 0.05

0.05 - (0.0031 x 5 {since it's a 5x dilution}) = 0.0345 moles of HCl reacted with the carbonate.

0.0345/2 (reaction goes 2:1 ratio), so 0.01725 moles of that MCO₃ were carbonate.

As you had 1.41g of MCO₃, 1.41/0.01725 = 81.73, which is the RMM of the MCO₃.

Carbonate molecular mass is C (12) + Ox3 (48) = 60. So the mass of M is 21.73. Magnesium is 24.3

Sorry if any of my calculations are wrong, it's late and I'm tired. Besides, it's so long since I've done this sort of stuff.

Marcus

Reply 2

Borek

marcusfox - indeed, you are wrong. Sr it is.

25 mL of the final solution contained 3.1 mmol of HCl, thus there was 31 mmol excess of HCl (25 mL of 250 mL was titrated - thus 10x).

Original amount of HCl used was 50 mmol. 50-31=19 mmol reacted with carbonate. There was 9.5 mmol of carbonate.

Molar mass of carbonate is 1.41/0.0095=148.4

So the molar mass of Me is 148.4-12-3*16=88.4

Sr is 87.6

Best,
Borek
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Reply 3

marcusfox