1. 2. 3. How many grams of lead(l) fluoride dissolve when 1.525 g of PbF2 is add
ID: 589100 • Letter: 1
Question
1. 2. 3. How many grams of lead(l) fluoride dissolve when 1.525 g of PbF2 is added to a 500.0 mL solution of 0.0125 M HNO, Determine the pH of a solution this is made by mixing 100.0 mL of a 0.250 M NaCIO, solution with 150.0 mL of a 0.375 M HCIO2 solution and 50.0 mL of a 0.205 M HCI solution. Below is the graph of the titration of 15.0 mL of acetic acid with NaOH. If (OH]- 0.155 M, what is the concentration of the acetic acid solution? Vinegar Titration Curve 14.00 13.00 12.00 11.00 10.00 9.00 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 0.00 .00 5.00 0.00 15.00 20.00 25.00 30.00 35.00 Volume NaOH (mL) 4. 5. Determine the molar solubility of HgCl, in water. Calculate the AG. for the dissolution of NH·NO-Explanation / Answer
1. moles of PbF2 present = 1.525 g/245.2 g/mol = 0.00622 mol
moles of F- present = 2 x 0.00622 mol = 0.01244 mol
moles of H+ added = 0.0125 M x 0.5 L = 0.00625 mol
so moles of PbF2 to dissolve = 0.00625 mol/2 = 0.003125 mol
mass of PbF2 to dissolve = 0.003125 mol x 245.2 g/mol = 0.77 g
2. initial moles of NaClO2 = 0.250 M x 100 ml = 25 mmol
initial moles of HClO2 = 0.375 M x 150 ml = 56.25 mmol
moles of HCl added = 0.205 M x 50 ml = 10.25 mmol
final moles of NaClO2 = 25 - 10.25 = 23.75 mmol
final moles of HClO2 = 56.25 + 10.25 = 66.50 mmol
pKa of HClO2 = 1.96
Using Hendersen-Hasselbalck equation,
pH = pKa + log(NaClO2/HClO2)
= 1.96 + log(23.75/66.50) = 1.513
3. From the graph, equivalence point volume of NaOH = 21 ml
moles of NaOH = 0.155 M x 21 ml = 3.255 mmol
moles of acetic acid reacted = 3.255 mmol
molarity of acetic acid solution = 3.255 mmol/15 ml = 0.217 M
4. Ksp of HgCl2 = 1.2 x 10^-18
Ksp = [Hg2+][Cl-]^2
molar solubility of HgCl2 in water,
1.2 x 10^-18 = (x)(2x)^2
molar solubility x = 6.7 x 10^-7 M
5. dGo for dissolution of NH4NO3
NH4NO3(s) --> NH4+(aq) + NO3-(aq)
dGo = dGo(products) - dGo(reactants)
= (-79 - 111) - (-184)
= -6 kJ/mol
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