I wrote the equations needed for each question but I don\'t exactly know how to
ID: 1011212 • Letter: I
Question
I wrote the equations needed for each question but I don't exactly know how to answer these. Please help! 196 EQULD400: Estimating the Solubiley Producz Constant of Seromtium lodate Using Microscale Techniqaes 3. A student used the techniques in this module to estimate the Kyp of lead iodide (Pbly). After preci- pitating and rinsing Pbly in a reaction pipet, the student equilibrated the solid in distilled water. The equilibrium reaction is shown in Equation 18 Pbl3(s) Pb(a) +21 (aq) (Eq. 18) The student analyzed 30 drops of the supernatant liquid for Pb- ion by titrating it with 28 drops of 1.23 × 10-'M EDTA, according to the reaction shown in Equation 19, Po(aq) + EDTA (aq) PbEDTA (a) that the reaction pipet delivered 33 drop mL" and the titration pipet (Eq. 19) The student determined that the reaction pipet delivered 33 drops ml delivered 25 drops ml.- ) Calculate the volume, in liters, ofone drop delivered from the reaction pipet,using Equation & (2) Calculate the volume, in liters, of equilibrated supernatant liquid titrated, using Equation 7 3) Calculate the volume,in liters, of one drop delivered from the titration pipet, using Equation 6
Explanation / Answer
1. 33 drops/ml is delievered from reaction pipet. So,
volume of 1 drop ml = 1 ml/33 drops = 0.0303 ml = 3.03*10-5 L
2. Supernatant liquid added = 30 drops.
volume of 1 drop = 3.03*10-5 L
So, volume of 30 drops = 30*3.03*10-5 L = 90.9 *10-5 L
3. 25 drops/ml is delievered from titration pipet. So,
volume of 1 drop ml = 1 ml/25 drops = 0.04 ml = 4.0*10-5 L
4. EDTA added = 28 drops.
volume of 1 drop = 4.0*10-5 L
So, volume of 28 drops = 28*4.0*10-5 L = 112.0 *10-5 L
5. Molarity of EDTA = 1.23*10-3 M
Volume of EDTA = 112.0 *10-5 L
Moles of EDTA = concentration*volume = 1.23*10-3 M*112.0 *10-5 L = 1.38*10-6 mol
6. According to equation 19, 1 mol of Pb2+ reacts with 1 mol of EDTA.
So, number of moles of Pb2+ = number of moles of EDTA = 1.38*10-6 mol
7. Volume of Pb2+ used = 90.9 *10-5 L
number of moles of Pb2+ = 1.38*10-6 mol
Concentration of Pb2+ = number of moles/volume
= 1.38*10-6 mol/90.9 *10-5 L = 0.00152 M = 1.52*10-3 M
8. In equation 18,
2 moles of I- are reacting with 1 mol of Pb2+.
So, concentration of I- = (1.52*10-3 M)*(2/1) = 3.04*10-3 M
9. Ksp of PbI2 = [Pb2+][I-]2
= (1.52*10-3 M)(3.04*10-3 M)2
= 14.05*10-9
Ksp Of PbI2 = 14.05*10-9