15.85 A 104.6 mL sample of 0.100 M methylamine (CH3NH2, K b=3.7×104) is titrated
ID: 1028493 • Letter: 1
Question
15.85 A 104.6 mL sample of 0.100 M methylamine (CH3NH2, Kb=3.7×104) is titrated with 0.265 M HNO3. Calculate the pH after the addition of each of the following volumes of acid.
A. 0.0 mL B. 19.7 mL C. 39.5 mL D. 59.2 mL
15.95 A. Use the following solubility data to calculate a value of Ksp for CdCO3: 1.0×106M
B. Use the following solubility data to calculate a value of Ksp for Ca(OH)2: 1.06×102M
C. Use the following solubility data to calculate a value of Ksp for PbBr2: 4.34 g/L
D. Use the following solubility data to calculate a value of Ksp for BaCrO4: 2.8×103 g/L
15.125 Consider saturated solution of the slightly soluble salt AgBr. Is the solubility of AgBr increased, decreased, or unaffected by the addition of each of the following substances?
A. When HBr is added the solubility of AgBr will
D. When NH3 is added the solubility of AgBr will
F. When Ba(NO3)2 is added the solubility of BaCO3 will
H. When CH3CO2H is added the solubility of BaCO3 will
Explanation / Answer
15.85
A. 0.0 ml
[CH3NH2] = 0.1 M
CH3NH2 + H2O <==> CH3NH3+ + OH-
let x amount reacted
Kb = [CH3NH3+][OH-]/[CH3NH2]
3.7 x 10^-4 = x^2/0.1
x = [OH-] = 6.08 x 10^-3 M
pOH = -log[OH-] = 2.21
pH = 14 - pOH = 11.79
B. 19.7 ml
initial CH3NH2 = 0.1 M x 104.6 ml = 10.46 mmol
added HNO3 = 0.266 M x 19.7 ml = 5.2402 mmol
moles CH3NH3+ formed = 5.2402 mmol
moles CH3NH2 remained = 5.2198 mmol
pH = pKa + log(CH3NH2/CH3NH3+)
= 10.57 + log(5.2198/5.2402) = 10.57
C. 39.5 ml
initial CH3NH2 = 0.1 M x 104.6 ml = 10.46 mmol
added HNO3 = 0.266 M x 39.5 ml = 10.507 mmol
excess [HNO3] = [H+] = 0.047 mmol/144.1 ml = 3.26 x 10^-4 M
pH = -log(3.26 x 10^-4) = 3.49
D. 59.2 ml
initial CH3NH2 = 0.1 M x 104.6 ml = 10.46 mmol
added HNO3 = 0.266 M x 59.2 ml = 15.7472 mmol
excess [HNO3] = [H+] = 5.2872 mmol/163.8 ml = 0.0323 M
pH = -log(0.0323) = 1.49
15.95
A. solubility = 1 x 10^-6 M
Ksp CdCO3 = [Cd2+][CO3^2-]
= (1 x 10^-6)^2
= 1 x 10^-12
B. solubility = 1.06 x 10^-2 M
Ksp Ca(OH)2 = [Ca2+][OH-]^2
= (1.06 x 10^-2)(2 x 1.06 x 10^-2)^2
= 4.76 x 10^-6
C. solubility = 4.34 g/L/367.01 g/mol
= 0.012 M
Ksp PbBr2 = [Pb2+][Br-]^2
= (0.012)(2 x 0.012)^2
= 6.91 x 10^-6
D. solubility = 2.8 x 10^-3 g/L/253.4 g/mol
= 1.10 x 10^-5 M
Ksp BaCrO4 = [Ba2+][CrO4^2-]
= (1.10 x 10^-5)^2
= 1.21 x 10^-10
15.125
Consider saturated solution of the slightly soluble salt AgBr. Effect on solubility of AgBr by,
A. When HBr is added the solubility of AgBr will decrease
D. When NH3 is added the solubility of AgBr will increase
F. When Ba(NO3)2 is added the solubility of BaCO3 will unaffected
H. When CH3CO2H is added the solubility of BaCO3 will increase
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