I2 + 2Na2(S2O3) → Na2(S4O6) + 2NaI
(28.68ml)(0.015 mmol/ml) = 0.4303 mmol [S2O3]2– consumed
Therefore, there were 0.4303/2 mmol = 0.2151 mmol I2 present
initial 0.280g HI = 280mg/(127.91 mg/mmol) = 2.189 mmol
.....2HI(g) ⇌ H2(g) + I2(g)
2.189-2x........x.........x
Therefore, final HI = 2.189 -2(0.2151) mmol
Kc = (0.2151)^2/(2.189 - 2(0.2151))^2 = 0.01496
Because the # of moles of reactants and products are equal, the volume of the bulb isn't needed (it cancels out of Kc). For the same reason Kc = Kp
