(Accidentally posted this question twice. ignore this)
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Acid titration question watch
- Thread Starter
Last edited by AndiP4ndi; 06-03-2018 at 17:21.
- 06-03-2018 16:51
(Original post by AndiP4ndi)
- 06-03-2018 17:13
I'm stuck on this practice question for A-level Chemistry:
An indigestion tablet containing sodium hydrogencarbonate was dissolved in 250cm^3 of water. 25.0cm^3 of this solution was found to need 16.50cm^3 of a 0.150mol dm^-3 solution of HCl for neutralisation.
NaHCO3+HCl --> NaCl + H2O + CO2
If the original mass of the indigestion tablet was 3.20g, calculate the percentage by mass of sodium hydrogencarbonate in the tablet.
I don't know how to work this out as I have never done a calculation like this before. All help is appreciated.
"Here is the worked solution which gives you % mass of 32.5%.
1. n(H+) titration = (16.5/1000) x 0.15 = 0.002475mol
work out the moles of H+/HCl needed to neutralise the HCO3- ion in the titration.
2. n(HCO3-) = 0.002475 / 2 = 0.0012375mol
look at the mole ratio in the reaction you provided. For every one HCO3- ion there were 2 H+ ions reacting. So we divide the moles of H+ by 2 to get moles of HCO3- in the titration.
3. n(HCO3-) total = 0.0012375 x 10 = 0.012375mol
remember that the mol of HCO3- calculated in step 2 was only in 25cm3 solution. Originally the tablet was dissolved in 250cm3, which is 10 times the volume. So the total amount of HCO3- is 10 times the amount in the titration. (i keep mentioning HCO3- because the NaHCO3 is ionic and will dissociate when in the solution)
4. mass of NaHCO3 = 0.012375 x (23 + 1 + 12 + 48) = 1.0395 g
now we know moles of NaHCO3 ( moles of HCO3- = moles of Na+ in the ionic lattice) and its Mr, we work out its mass
5. % mass = (1.0395 / 3.2) x 100 = 32.484375% which rounds to 32.5% "Last edited by dip0; 06-03-2018 at 18:04.
- 06-03-2018 17:51
Ignore the above solution its wrong. The mole ratio is 1:1 not 1:2.