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A sodium carbonate solution was prepared by dissolving 10.6g sodium carbonate in 100cm3 water...?
...was titrated against dilute hydrochloric acid. 26.6cm3 of acid was required for the neutralisation. Is the question, I need the answers to these questions:
1. Write a balanced equation for the reaction
2. Calculate the moles of sodium carbonate in stock solution
3. Calculate moles of sodium carbonate in 25cm3 sample
4. Calculate moles of hydrochloric acid used in titration
5. Calculate concentration of the hydrochloric acid
Thanks!
4 Answers
- 1 decade agoFavorite Answer
1. Na2CO3 + HCl -> NaCl + NaHCO3
NaHCO3 is sodium hydrogen carbonate
2. Moles = mass/RMM
moles of Na2CO3 =10.6/106 = 0.1
3. 0.1/4 = 0.0025 moles of Na2CO3 in 25 cm3
4. 1:1 ratio, so 0.0025 moles of HCl reacted
5. Concentration = moles/(volume/1000)
[HCl] = 0.0025/0.0266 = 0.094 2d.p.
- MargaretLv 45 years ago
10 mL of carbonate took 18.2 cm3 of 0.1 M HCl which is (0.1/1000) x 18.2 moles = 0.00182 moles acid used reaction is 2 HCl to 1 carbonate, so moles of carbonate = moles HCl/2 = 0.00182/2 = 0.00091 moles carbonate in 10 cm3 (moles used in the titration) which would give 0.091 moles carbonate in a litre or 0.091 M
- Anonymous1 decade ago
Hmm, methinks someone has double chemistry first thing monday morning.
He he, nice try but no cigar!
- ray dLv 41 decade ago
Do your own Chemistry Homework!
How Lazy?
No-one used to do my Chemistry homework for me, just do what I used to do. make it all up.
