2. Calculate the number of moles of sodium carbonate present in a mixture of sod

Question

2. 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 HCl were used to titrate 0.200 g of this mixture with phenolphthalein as the indicator. 3 Calculate the mass of sodium carbonate present in the mixture in question 2. 4. 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 HCI were used in step 2 of the titration, using methyl orange as the indicator.

Explanation / Answer

2 ) Assuming that 6 mL HCl is consumed by both Na2CO3 & NaHCO3:

Consider the following equations

Na2CO3 + 2HCl ---------> 2NaCl + CO2 + H2O

NaHCO3 + HCl ---------> NaCl + CO2 + H2

The above equ(s) state that Na2CO3 & NaHCO3 consume HCl in the ratio of 2 : 1

Now if entire mixture consumes 6 mL, it should be resolved into 2 : 1 ratio

Then it becomes 4 : 2

ie Na2CO3 consumed 4 mL of HCl
& NaHCO3 consumed 2 mL of HCl
Remaining Neutral Compound does not interfere in the titration

Now comes determination of:
1) Amt of Na2CO3
During titration
Na2CO3 + 2HCl ---------> 2NaCl + CO2 + H2O

Let M = n* 1000/V (Vol in mL)
so MV =n * 1000 (where n = moles of Na2CO3 present in the mixture)

M1V1/n1 = M2V2/n2

Of Na2CO3
M1V1 = n *1000
n1 = 1 mol (moles of Na2CO3 consumed according to the balanced equ)

Of HCl
M2 = 0.100 M
V2 = 4 mL (see at resolved ratio)
n2 = 2 mol

substituting values
(n * 1000)/1 = (0.1 * 4)/2
n = (0.1 * 4)/(2 * 1000)
n = 0.0002 mol

3 ) Then the amt of Na2CO3 present is = n * Molecular weight
= 0.0002 * 106
= 0.0212 g

4) Determine the amt of NaHCO3 from the following equ

NaHCO3 + HCl ---------> NaCl + CO2 + H2O

In this case NaHCO3 consumes 5 mL of HCl in titration
n*1000 = 0.1*5 , n = 0.0005 mol
Finally
no of moles of NaHCO3 = 0.0005 mol
and the amt of NaHCo3 = 0.042 g

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