2005 HSC Chemistry question :p (1 Viewer)

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Question 24 ​

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(5 marks)
An antacid tablet is known to contain calcium carbonate (CaCO3). To determine the
mass of calcium carbonate in the tablet, the following procedure was used.
• The tablet was crushed and then placed in a beaker.

• A pipette was used to add 25.0 mL of 0.600 mol L–1 hydrochloric acid to the
crushed tablet in the beaker.

• Once the reaction between the calcium carbonate and hydrochloric acid had
stopped, phenolphthalein indicator was added to the reaction mixture.

• A teflon-coated burette was then used to add 0.100 mol L–1 sodium
hydroxide to the beaker to neutralise the excess hydrochloric acid.

• The phenolphthalein changed from colourless to pink after 14.2 mL of the
sodium hydroxide solution had been added.

(c) Calculate the mass of calcium carbonate in the original antacid tablet. (3)
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could someone show me how to do this?

 

[what else?]

anyone?

 

[minijumbuk]

Best way to do these are to do it backwards from the information they gave.

HCl + NaOH --> NaCl + H₂O

So the number of moles of NaOH that reacted to neutralise the HCl would be:
0.1 x 0.0142 = 1.42x10^-3 mol

Therefore 1.42x10^-3 mol of HCl was excess

Initially, 0.025 x 0.6 = 0.015 mol was poured into antacid tablet
Taking away the excess moles, we can find out how many moles actually reacted with CaCO₃
Therefore no. of moles of HCl that reacted with CaCO₃ = 0.015 - 1.42x10^-3
= 0.01358 mol of HCl.

CaCO₃ + 2HCl --> CaCl₂ + CO₂ + H₂O

Since CaCO₃:HCl = 1:2, the no. of moles of CaCO₃ which reacted = 0.01358 / 2
= 6.79x10^-3 mol of CaCO₃

Therefore there was 6.79x10^-3 mol of CaCO₃ initially.

Mass of CaCO₃= 6.79x10^-3 x (40.08 + 12.01 + 3x16)
=0.68 g (2 sig. fig.)

 

[fozzster22]

i just so happened to be doing this question too actually

first off let's make sure we got the right answer for a and b

a: equation was CaCO3 + 2HCl --> CO2 + H2O + CaCl2

b: moles of HCl was .6*.025 = .015 mol added

now for part c, we gotta work backwards sort of. how much NaOH reacted? same amount as HCl, if you use this equation: NaOH+HCl-->H2O + NaCl, and assuming that they titrated perfectly (of course they did!)

now according the data given, 14.2 ml of .1 mol/L NaOH was added, so

NaOH reacted: .1*.0142 = 1.42*10^-3 mol

how much HCl is reacted? same amount as NaOH, 1.42*10^-3 mol

the tricky part..

how much HCl reacted with CaCO3? it's the difference of the total amount of HCl added (.015) and the amount of which reacted with the NaOH (1.42*10^-3),

.015 - 1.42*10^-3 = .01358 mol

(we got the .015 from part b.)

this divided by 2 since ratio HCl:CaCO3 is 2:1, according to that other equation we got in part a)

n mol CaCO3 reacted = 6.79*10^-3

then multiply by the molar mass..

6.79*10^-3 * 100.09 = .684432 g

hope i helped

 

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cheers

 

[henry08]

Yeah I solved this a few days ago and had trouble with it initally as well. Teacher said it was a hard question though.

 

[minijumbuk]

It's actually quite a typical calculation question for such value of marks.