What are the bad effects of sodium carbonate?

25.0cm(3) of a solution of sodium carbonate reacted with exactly 20.0cm(3) of 0.2mol per dm(3) of nitric acid.?

• The equation for the reaction is: Na2CO3 + 2HNO3 --> 2NaNO3 + CO2 + H20 a) name a suitable indicator that could be used to find the end point of this reaction (is this phenolphthalein?) b) calculate the number of moles of nitric acid in 20.0cm(3) of the solution c) calculate the number of moles of sodium carbonate in 25.0cm(3) of solution d) calculate the concentration of sodium carbonate solution in mol per dm(3) e) calculate the concentration of the solution in grams per dm(3) of anhydrous sodium carbonate Ar values: Na= 23, C=12, O=16 (3) means cubed

• Answer:

  a. I think this is actually the trickiest of the five parts. There are at least three factors to take into consideration when selecting an indicator for this reaction. Firstly, are you adding the Na2CO3 to the HNO3, or vice versa? If you’re adding the Na2CO3, then phenolphthalein might seem suitable. It’s easy to detect the first pink coloration, which occurs at a pH between 8 and 9. If you’re adding the acid to the base, it’s not so easy to decide when the pink colour disappears. Another thing to take into account is the fact that you’re releasing CO2. This will dissolve partially in the solution, forming carbonic acid, H2CO3. So, even when you’ve reacted all the Na2CO3 and HNO3, the mixture will still be slightly acidic. So you might be better off with an indicator that will change colour in an acid environment. As acidity decreases, methyl orange changes from red to orange in a pH range of about 4 to 5. Thirdly, the whole issue of titrating a weak base with a strong acid is complicated: the equivalence point doesn’t occur at pH = 7, but at some lower pH. --------------------------------------... In questions relating to molar concentrations, you use the definition of molar concentration. If n is the number of moles of solute, and V is the volume of solution in L (or dm^3), then the molar concentration c is … c = n / V which rearranges to give … n = cV --------------------------------------... b. For the nitric acid solution: V = 20.0 mL = 0.0200 L c = 0. 200 mol / L (I’ve used 3 significant digits for c) n = cV = (0. 200 mol / L) ( 0.0200 L) = 0. 00400 mol --------------------------------------... c. Na2CO3 + 2HNO3 --> 2NaNO3 + CO2 + H20 This equation tells you that 1 mole of Na2CO3 reacts with 2 moles of HNO3. So, if you had 0.00400 mol of HNO3, you must have had half as much Na2CO3, i.e. 0.00200 mol. --------------------------------------... d. For the Na2CO3 solution, n = 0.00200 mol V = 25.0 mL = 0.0250 L c = n / V = 0.00200 mol / 0.0250 L = 0.0800 mol/L --------------------------------------... e. To convert moles to grams, use … mass = moles x molar mass For Na2CO3, the molar mass is (2 x 23) + 12 + (3 x 16), which comes to 106 g/mol. And 1 L of your solution contains 0.080 mol of Na2CO3. So the mass, m, of dissolved Na2CO3 is … m = (0.0800 mol) (106 g/mol) = 8.48 g --------------------------------------... I hope this is of some use to you. Check my figures to be sure I haven’t made any mistakes.