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AS Chemistry titration hard question - could someone please help (Working Included)
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A sample containing ammonium sulphate was warmed with 250 cm3 of 0.800M sodium hydroxide solution. After the evolution of ammonia had ceased, the excess of sodium hydroxide solution was neutralised by 85.0 cm3 of hydrochloric acid of molarity 0.500M. What mass of ammonium sulphate did the sample contain?(
a) (NH4)2SO4 + 2NaOH → Na2SO4 + 2H2O + 2NH3
(b) NaOH + HCl → NaCl + H2O
Answer so far: 2NaOH = 0.2moles, 0.25dm3, and 0.8M, therefor the Ammonium Suplhate is 0.1moles
I have worked out that 13.21g Ammonium Sulphate was used: Because the RMM of Ammonium Suplhate is 132.1, therefore because there is 0.1 moles, 0.1 x 132.2 = 13.21g.
I have also worked out that 16g of NaOH was used: Because the RMM of NaOH is 40, therefore because there is 0.2 moles, 0.2 x 40 = 8, and because there are 2 moles in the organ all formula, 8 x 2 = 16.
The HCl is 0.0425 moles, 0.085dm3 and 0.5M, therefor the NaOH is 0.0425moles. From the I have worked out that 1.7g of NaOH and 1.55g of HCl were used.
Could someone plazas help me with the rest as I can't get an answer?
a) (NH4)2SO4 + 2NaOH → Na2SO4 + 2H2O + 2NH3
(b) NaOH + HCl → NaCl + H2O
Answer so far: 2NaOH = 0.2moles, 0.25dm3, and 0.8M, therefor the Ammonium Suplhate is 0.1moles
I have worked out that 13.21g Ammonium Sulphate was used: Because the RMM of Ammonium Suplhate is 132.1, therefore because there is 0.1 moles, 0.1 x 132.2 = 13.21g.
I have also worked out that 16g of NaOH was used: Because the RMM of NaOH is 40, therefore because there is 0.2 moles, 0.2 x 40 = 8, and because there are 2 moles in the organ all formula, 8 x 2 = 16.
The HCl is 0.0425 moles, 0.085dm3 and 0.5M, therefor the NaOH is 0.0425moles. From the I have worked out that 1.7g of NaOH and 1.55g of HCl were used.
Could someone plazas help me with the rest as I can't get an answer?
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#3
There were indeed 0.200 mol of NaOH at the start
There were indeed 0.0425 mol of HCl needed to react with the excess NaOH
Which mean the (NH4)2SO4 reacted with the rest of the NaOH, i.e. 0.1575 mol of NaOH reacted with the (NH4)2SO4
The amount of (NH4)2SO4 must have been 0.1575/2 = 0.07875
When multiplied by the Mr of (NH4)2SO4 you get 10.4.
But it is late and I have drunk much.
There were indeed 0.0425 mol of HCl needed to react with the excess NaOH
Which mean the (NH4)2SO4 reacted with the rest of the NaOH, i.e. 0.1575 mol of NaOH reacted with the (NH4)2SO4
The amount of (NH4)2SO4 must have been 0.1575/2 = 0.07875
When multiplied by the Mr of (NH4)2SO4 you get 10.4.
But it is late and I have drunk much.
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