A 0.4510 g metal sample is dissolved in nitric acid and all metals are masked except bismuth (Bi^3+ = 209 g/mol). The solution is adjusted to pH 10 and titrated with 0.0150M 21.15ml EDTA. What is the percentage of bismuth in the sample?
Na2EmillimDTA form a complex with Bi^3+ in a 1:1 ratio.
moles EDTA = L x M = 0.02115 L x 0.0150 M = ?
moles Bi^+ must be the same since it is a 1:1 ratio.
g Bi metal = mols Bi from above x 209 g/mol = ?
Then %Bi metal = (grams Bi/mass sample)*100 = ?