A 0.1105 g sample of primary standard grade Na2CO3 (M.M. = 105.99) was dissolved
ID: 773983 • Letter: A
Question
A 0.1105 g sample of primary standard grade Na2CO3 (M.M. = 105.99) was dissolved in 55 mL of distilled H2O. The solution required 22.18 mL of the HCl solution to reach the methyl orange end point. a. Calculate the number of mmoles of primary standard grade Na2CO3. b. How many mmoles of HCl reacted with the Na2CO3? c. Calculate the molarity of the HCl solution.
A 0.1105 g sample of primary standard grade Na2CO3 (M.M. = 105.99) was dissolved in 55 mL of distilled H2O. The solution required 22.18 mL of the HCl solution to reach the methyl orange end point. a. Calculate the number of mmoles of primary standard grade Na2CO3. b. How many mmoles of HCl reacted with the Na2CO3? c. Calculate the molarity of the HCl solution.
Explanation / Answer
Answer:-
Na2CO3(s) + 2HCl(aq) ---> 2NaCl(aq) + H2O(l) + CO2(g)
Boiling will remove the CO2 gas, which can make the solution more acidic through the equilibrium:
CO2 + H2O <----> H2CO3 <----> H+ + HCO3-
now
1. First find out how many moles of NaHCO3 you have. Do this by figuring out the molecular mass:
1Na(23.0) + 1H(1.008) + 1C(12.0) + 3O(16.00) =84.008 g/mole
1.10g(1mole/84.008g)=0.0131 moles of NaHCO3
Let v = volume of HCl
Assume one mole of HCl neutralizes one mole of NaHCO3.
v(0.200moles/L)=0.0131moles
v=0.0131moles/0.200moles/L
v=0.0655L or 65.5mL
2. Do this one the same way as number 1.
3. Phenolphthalein endpoint is for the strong base NaOH.
23.98mL(1L/1000mL)(0.100moles/L)=0.002
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