b) n(Na2CO3) = 1.236/106.0 = 0.01166mol
c(Na2CO3) = 0.01166/0.250 = 0.04664 M
*I check if b) was right*
c)n(HCl) = 2 x n(Na2CO3) = 0.02332mol
c(HCl) = 0.02332/0.02099 = 1.111 M
It's c) that doesn't come off right. Am I doing a step incorrectly?
Thanks
"c)n(HCl) = 2 x n(Na
2CO
3) = 0.02332mol"
Your issue is here
the n(HCl) in the titre is indeed twice the amount of Na
2CO
3 in the aliquot. However, the amount of Na
2CO
3 in the aliquot is not 0.01166 mol, this is the amount of Na
2CO
3 in the 250.0 mL volumetric flask. Remember you are taking 20.00 mL aliquots of the Na
2CO
3 from the 250.0 mL sample.
Alternative solution:
n(Na
2CO
3)
250.0 mL = 1.236/106.0 = 0.01166mol
n(Na
2CO
3)
20.00 mL aliquot = 0.01166 x 20.00/250.0 = 0.0009328 mol
n(HCl)
titre = n(Na
2CO
3)
20.00 mL aliquot x 2 = 0.001866 mol
[HCl] = 0.001866/0.02099 = 0.08888 M
I bet you anything the 0.08892 M answer is because they rounded off too early.