A student was given a 0.10 M solution of an unknown diprotic acid H2A and asked to determine the Ka1 and Ka2 values for the diprotic acid. The student titrated 45.8 mL of the 0.10 M H2A with 0.10 M NaOH. After 22.9 mL of NaOH was added, the pH of the resulting solution was 6.70. After 45.8 mL of NaOH was added, the pH of the resulting solution was 8.00. What are the values of Ka1 and Ka2 for the diprotic acid?

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.