Question:
An unknown mass of (NH4)2CO3 is mixed with 4g of NaOH and dissolved in 100cm3 of water,which results a clear solution.Gasses are removed by heating the final solution.
2 parts of 25cm3 of final solution is taken and one is titrated with 0.5 M HCl and under Hph indicator and the other with the same HCl under Me.O indicator.
H
Burrete readings under the Hph and Me.O as indicators are respectively 40 and 50cm3.
Find the mass of the initial solid sample.
My thoughts on the question:
Reactions:
(NH4)2CO3+ 2NaOH---> Na2CO3+ 2NH4OH
Hph;
1)Na2CO3+2HCl--->NaHCO3+NaCl
Me.O;
1)Na2CO3+ 2HCl --->2NaCl+ CO2 +H2O
2)NH4OH+Hcl-->NH4Cl+H2O
I know that under Me.O as the indicator it shows colour change after both Na2CO3 and NH4OH neutralized.So by that we can find the moles of NH4OH,formed.And the moles of Na2CO3,formed,can be determined by the moles of HCl used under Hph as the indicator.
But how do we find the mass of the initial sample(Ammonium Carbonate)?
Is my approach correct or am I missing anything?
If we assume there was x moles of Ammonium Carbonate at the beginning,do we have the consider the moles reacted and etc.like I've shown in the below OR is there any wiser approach as this is an Multiple Choice question...
From my approach in 25cm3,
Final moles of Na2CO3=(0.5)*40*(10)^-3=2*(10)^-2
Final moles of NH4OH= (0.5)*10*(10)^-3= 0.5*(10)^-2
(NH4)2CO3+NaOH-->NH4OH+Na2CO3
Initial moles x 0.1 - -
Final moles ? ? 0.005 0.02
I assume Hph stands for phenolphthalein, or if not, then it changes color in the same pH range as phenolphthalein which is approx 8.3 or so. Also, I assume that all of the (NH4)2CO3 reacts, all of the NH3 is boiled away, and an excess of NaOH remains with the Na2CO3. Your equation 1 with Hph and 2 with MeO are incorrect. Here is what happens in the two titrations.
In the Hph titration you titrate all of the OH^- there + the CO3^2- half way (CO3^2- is titrated to HCO3^-).
So you have 40 mL x 0.5M = 20 mmols OH + CO3^2- half way.
For the MeO titration you titrate all of OH^- AND ALL of the CO3^2-. That was 50 mL x 0.5M = 25 mmols. The volume required for the OH^- is the same in both cases so the difference (25 mmols-20 mmols = 5 mmols) must be for titrating the CO3^2- to HCO3^- which would be 10mmols for titrating all of the CO3^2- to CO2 and H2O. That tells you that you had 20-5 = 15 mmols OH^- (or 25-10 = 15 mmols OH^-) mmols OH^- titrated.That 15 mmols was in a 25 mL aliquot sample of 100 mL so the sample contained 60 mmols OH^-. That is 60 mmols NaOH left after the reaction. You had 4.0g NaOH or 0.1 mol (100 mmols) so the reactin used 100-60 = 40 mmols). Use stoichiometry to convert to mols (NH4)2CO3 and convert to grams.
Hope this helps.
Thank you for the explanation
Your approach is almost correct. To find the mass of the initial solid sample (ammonium carbonate), you need to consider the moles of both Na2CO3 and NH4OH formed in the reactions.
Let's start by calculating the moles of Na2CO3 formed using the Hph titration:
Moles of Na2CO3 = Moles of HCl used under Hph indicator
Moles of HCl used under Hph indicator = 40 cm3 * 0.5 M HCl = 20 mmol (millimoles)
Since the balanced equation for the reaction between Na2CO3 and HCl is:
Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O
We can see that for every 1 mole of Na2CO3, 2 moles of HCl react. Therefore, the moles of Na2CO3 formed is also 20 mmol.
Next, let's calculate the moles of NH4OH formed using the Me.O titration:
Moles of NH4OH = Moles of HCl used under Me.O indicator
Moles of HCl used under Me.O indicator = 50 cm3 * 0.5 M HCl = 25 mmol (millimoles)
Since the balanced equation for the reaction between NH4OH and HCl is:
NH4OH + HCl -> NH4Cl + H2O
We can see that for every 1 mole of NH4OH, 1 mole of HCl reacts. Therefore, the moles of NH4OH formed is also 25 mmol.
Now, let's consider the reaction between (NH4)2CO3 and NaOH:
(NH4)2CO3 + 2NaOH -> 2NH4OH + Na2CO3
From the balanced equation, we can see that for every 1 mole of (NH4)2CO3, 2 moles of NH4OH are formed. Therefore, the moles of (NH4)2CO3 initially present can be calculated as follows:
Moles of (NH4)2CO3 = (Moles of NH4OH) / 2 = 25 mmol / 2 = 12.5 mmol
Finally, to calculate the mass of the initial solid sample (ammonium carbonate), we use the molar mass of (NH4)2CO3:
Molar mass of (NH4)2CO3 = 2 * (14 + 1) + 12 + 3 * 16 = 96 g/mol
Mass of initial solid sample = Moles of (NH4)2CO3 * Molar mass of (NH4)2CO3
= 12.5 mmol * 96 g/mol
= 1.2 g
Therefore, the mass of the initial solid sample (ammonium carbonate) is 1.2 grams.