Please answer all 3 parts and show steps. Thanks! Calculate the number of moles
ID: 536760 • Letter: P
Question
Please answer all 3 parts and show steps. Thanks!
Calculate the number of moles of sodium carbonate present in a mixture of sodium bicarbonate, sodium carbonate and a neutral component if 6.0 ml of 0.100 M HCI were used to titrate 0.200 g of this mixture with phenolphthalein as the indicator. calculate the mass of sodium carbonate present in the mixture in question 2. Calculate the number of moles of sodium bicarbonate originally present in the unknown mixture in question 2, knowing that 15.0 ml of 0.100 M HCl were used in step 2 of the titration, using methyl orange as the indicator.Explanation / Answer
2) There are two points to remember:
a) Phenolphthalein will titrate sodium carbonate, Na2CO3 to NaHCO3. Phenolphthalein will not titrate NaHCO3 further to the final end point. Therefore, when phenolphthalein is used, Na2CO3 is half titrated.
b) Methyl orange will titrate NaHCO3 to produce NaCl, CO2 and H2O.
The titration of Na2CO3 with HCl using phenolphthalein indicator is as below:
Na2CO3 + HCl ------> NaHCO3 + NaCl …….(1)
As per the stoichiometric equation above,
1 mole Na2CO3 = 1 mole HCl = 1 mole NaHCO3
Moles of HCl used = (volume of HCl in L)*(concentration of HCl in mol/L) = (6.00 mL)*(1 L/1000 mL)*(0.100 mol/L) = 0.00060 mole (1 M = 1 mol/L)
Therefore, moles of Na2CO3 neutralized using phenolphthalein indicator = 0.00060 mole. This corresponds to neutralization of half the amount of Na2CO3 present in the sample. Hence, the number of moles of Na2CO3 present in the sample will be (2*0.00060) mole = 0.0012 mole (ans).
3) Molar mass of Na2CO3 = (2*22.9897 + 1*12.01 + 3*15.9994) g/mol = 105.9876 g.
Mass of Na2CO3 present in the sample = (0.0012 mole)*(105.9876 g/mol) = 0.12718 g 0.1272 g (ans).
4) 0.00060 mole Na2CO3 in part (2) required 6.00 mL of 0.100 M HCl; therefore, 0.0012 mole of Na2CO3 will require (2*6.00) mL = 12.00 mL of 0.100 M HCl to titrate to NaCl.
However, the total volume of HCl required to titrate NaHCO3 to NaCl using methyl orange indicator is 15.00 mL, as given. This equals the sum of NaHCO3 obtained from Na2CO3 (using phenolphthalein) and NaHCO3 present in the original sample.
Therefore, volume of HCl required to titrate the NaHCO3 present originally in the sample = (15.00 – 12.00) mL = 3.00 mL.
Moles of HCl required to titrate only NaHCO3 present in the original sample = (3.00 mL)*(1 L/1000 mL)*(0.100 mol/L) = 0.00030 mole.
The titration reaction is
NaHCO3 + HCl -------> NaCl + CO2 + H2O
As per the stoichiometric equation,
1 mole HCl = 1 mole NaHCO3.
Therefore, moles of NaHCO3 present in the original mixture = 0.00030 mole (ans).
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