chem(mole)

2007-02-22 8:04 pm

when5.72g of hydrated sodium carbonate(Na2CO3.nH2O)is dissolved in water and titrate with 1.0M sulphuric acid.20 cm^3 of sulphuric acid is needed for complete reaction.What is the value of n?

when 1.0g of an impure sodium carbonate sample is dissolved in water and titrate with 0.5M HCl,15.1cm^3 of HCl is needed.Calculate the percentage purity of sodium carbonate in the sample

更新1:

when Vcm^3 of 0.2M sulphuric acid is mixed with Vcm^3 of 0.1M hydrochloric acid and titrate with 0.2M sodium hydroxide. Calculate the volume of sodium hydroxide needed. (represent your answer in terms of V)

更新2:

回答 (1)

[匿名]

2007-02-22 9:02 pm

✔ 最佳答案

1. no of mole of sulphuric acid used = 1.0 x 20 x 10^-3 = 0.02 mol
By balance equation, Na2CO3‧nH2O + H2SO4 →Na2SO4 + CO2 + (n+1) H2O
no of mole of Na2CO3‧nH2O reacted = 0.02 mol
so 0.02mol of Na2CO3‧nH2O is 5.72g
i.e. its molecular mass = 5.72 / 0.02 = 286 g mol^-1
the formula mass of Na2CO3‧nH2O = 2x23+12+3x16+nx(2x1+16)
= 106+18n
so 106+18n = 286 , 18n = 180 , n = 10 //

2. no of mol of HCl used = 0.5 x 15.1 x 10^-3 = 7.55 x 10^-3 mol
By balance equation , Na2CO3 + 2HCl → 2NaCl + CO2 + H2O
no of mol of Na2CO3 reacted = 7.55 x 10^-3 / 2 = 3.775 x 10^-3 mol
so there are 3.775 x 10^-3 mol pure Na2CO3
since the molar mass of Na2CO3 = 2x23+12+3x16 = 106g mol^-1
mass of pure Na2CO3 = 106 x 3.775 x 10^-3 = 0.40015g
Hence the percentage purity of Na2CO3 = 0.40015 / 1.0 x 100% = 40.015% //

3. no of mol of hydrogen ion in H2SO4 = 0.2 x V x 10^-3 = 2V x 10^-4 mol
no of mol of hydrogen ion in HCl = 0.1 x V x 10^-3 = V x 10^-4 mol
so there are 3V x 10^-4 mol hydrogen ion in 2V cm^3 i.e. 0.15M
By balance equation , hydrogen ion + hydroxide ion → water
no of mol of hydroxide ion needed = 3V x 10^-4 mol
volume of NaOH needed = 3V x 10^-4 / 0.2 = 1.5V x 10^-3 dm^3 i.e. 1.5V cm^3 //

參考: By myself

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