Chem 55 Exam 2 Solutions

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CHEM 55 Fall 2007 Exam 1 Copy # 1 Name: Au,Hai Nguyen 3. Calculate the molarity of La(IO 3 ) 3 at saturation in water (K SP (La(IO 3 ) 3 = 1.0x10 -11 ). Ksp ⎛ ⎜ ⎝ x⋅ ( 3x) 3 27x 4 x 1.0⋅ 10 − 11 27 ⎞ ⎟ ⎠ 1 4 ⎛ ⎜ ⎝ Ksp 27 ⎞ ⎟⎠ 1 4 = 7.801× 10 − 4 = [La+3] 5 4. Calculate the molarity of La(IO 3 ) 3 at saturation in water containing 0.050 M LiIO 3 given K SP (LaIO 3 ) = 1.0x10 -11 . Ksp x⋅ 0.050 3 1.0⋅ 10 − 11 = 8× 10 − 8 = [La+3] 0.050 3 5 5. Repeat the above calculation but including the effects of ion activity and given that the ionic radius of La +3 is 900 pm and that of IO 3 - is 450 pm. 5 − 0.51⋅z 2 ⋅ μ 1+ α⋅ μ 305 z:= 3 α := 900 γLa := 10 γLa = 0.24 − 0.51⋅z 2 ⋅ μ 1+ α⋅ μ 305 z:= −1 α := 405 γIO3 := 10 γIO3 = 0.82 Ksp x⋅ γLa⋅0.050 3 ⋅γIO3 3 1.0⋅ 10 − 11 0.050 3 ⋅γIO3 3 ⋅γLa = 6.1 × 10 − 7 = [La+3]
CHEM 55 Fall 2007 Exam 1 Copy # 1 Name: Au,Hai Nguyen 6. A solution of NaOH was standardized by titration of a known quantity of the primary standard potassium hydrogen phthalate (KHP, FM 204.221 g/mol). a. Titration of 0.824 g of potassium hydrogen phthalate required 38.314 mL of NaOH to reach the endpoint detected by phenolphthalein. Find the concentration of NaOH in molarity (M). 10 mol⋅ KHP 0.824⋅ g⋅ ⋅ 204.221g ⋅ ⋅KHP 1⋅ NaOH 1⋅KHP ⋅ 1 = 38.31410 ⋅ − 3 ⋅L⋅ NaOH mol NaOH 0.105 ⋅ L b. A 100.00 mL aliquot of H 2 SO 4 solution required 57.911 mL of the NaOH solution to reach the endpoint. Calculate the concentration of the H 2 SO 4 . 5 57.91110 ⋅ − 3 0.1053mol ⋅ ⋅ NaOH 1⋅H2SO4 1 ⋅L⋅ NaOH⋅ ⋅ ⋅ = L⋅ NaOH 2⋅ NaOH 100.0⋅ 10 − 3 ⋅L⋅ H2SO4 mol H2SO4 0.0305 ⋅ L
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Chem 55 Exam 2 Solutions

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