1. To calculate the Ksp for silver bromide, we need to set up the equilibrium expression and solve for Ksp.
AgBr (s) ⇌ Ag+ (aq) + Br- (aq)
Ksp = [Ag+][Br-]
Since the solubility of silver bromide is given as 8.8 x 10^-7 mol/L, we can assume that the concentration of Ag+ and Br- ions is also 8.8 x 10^-7 mol/L.
Therefore, Ksp = (8.8 x 10^-7)(8.8 x 10^-7) = 7.74 x 10^-13
2. To calculate the Ksp for iron (II) hydroxide, we need to determine the concentration of Fe2+ and OH- ions using the solubility information.
Fe(OH)2 (s) ⇌ Fe2+ (aq) + 2OH- (aq)
The solubility is given as 1.4 x 10^-3 g/L, so we need to convert this to mol/L using the molar mass of Fe(OH)2. The molar mass of Fe(OH)2 is 89.85 g/mol.
1.4 x 10^-3 g/L * (1 mol/89.85 g) = 1.56 x 10^-5 mol/L
Since there is a 1:1 ratio between Fe2+ and OH- ions, the concentration of OH- ions is also 1.56 x 10^-5 mol/L.
Therefore, Ksp = (1.56 x 10^-5)(1.56 x 10^-5) = 2.43 x 10^-10
3. The solubility of copper (II) bromide is given as 2.0 x 10^-4 mol/L. Since copper (II) bromide is an ionic compound, it dissociates into ions in water.
CuBr2 (s) ⇌ Cu2+ (aq) + 2Br- (aq)
The concentration of Cu2+ ions is given as 2.0 x 10^-4 mol/L, and the concentration of Br- ions is 2 times that, so it is 4.0 x 10^-4 mol/L.
Therefore, Ksp = (2.0 x 10^-4)(4.0 x 10^-4) = 8.0 x 10^-8
4. The solubility of strontium sulfate is given as 5.83 x 10^-3 mol/L. Since strontium sulfate is an ionic compound, it dissociates into ions in water.
SrSO4 (s) ⇌ Sr2+ (aq) + SO4^2- (aq)
The concentration of Sr2+ ions is given as 5.83 x 10^-3 mol/L, and the concentration of SO4^2- ions is also 5.83 x 10^-3 mol/L.
Therefore, Ksp = (5.83 x 10^-3)(5.83 x 10^-3) = 3.40 x 10^-5
5. The solubility of magnesium hydroxide is given as 1.5 x 10^-4 mol in 100 mL. We need to convert this to mol/L by dividing by the volume in liters.
1.5 x 10^-4 mol / (100 mL * 0.001 L/mL) = 1.5 x 10^-3 mol/L
Since magnesium hydroxide dissociates into Mg2+ and 2OH- ions, the concentration of Mg2+ ions is 1.5 x 10^-3 mol/L, and the concentration of OH- ions is 2 times that, so it is 3.0 x 10^-3 mol/L.
Therefore, Ksp = (1.5 x 10^-3)(3.0 x 10^-3)(3.0 x 10^-3) = 1.35 x 10^-8
6. The solubility of silver sulfide is given as 1.3 x 10^-3 mol in 50 mL. We need to convert this to mol/L by dividing by the volume in liters.
1.3 x 10^-3 mol / (50 mL * 0.001 L/mL) = 2.6 x 10^-2 mol/L
Since silver sulfide dissociates into Ag+ and S2- ions, the concentration of Ag+ ions is 2.6 x 10^-2 mol/L, and the concentration of S2- ions is also 2.6 x 10^-2 mol/L.
Therefore, Ksp = (2.6 x 10^-2)(2.6 x 10^-2) = 6.76 x 10^-4
7. The solubility of barium fluoride is given as 2.2 x 10^-2 mol/L. Since barium fluoride is an ionic compound, it dissociates into ions in water.
BaF2 (s) ⇌ Ba2+ (aq) + 2F- (aq)
The concentration of Ba2+ ions is given as 2.2 x 10^-2 mol/L, and the concentration of F- ions is 2 times that, so it is 4.4 x 10^-2 mol/L.
Therefore, Ksp = (2.2 x 10^-2)(4.4 x 10^-2) = 9.68 x 10^-4
8. The solubility of lithium carbonate is given as 1.5 g/100 mL. We need to convert this to mol/L by dividing by the molar mass of lithium carbonate (73.89 g/mol) and the volume in liters.
1.5 g / (100 mL * 0.001 L/mL * 73.89 g/mol) = 2.03 x 10^-3 mol/L
Since lithium carbonate dissociates into Li+ and CO3^2- ions, the concentration of Li+ ions is 2.03 x 10^-3 mol/L, and the concentration of CO3^2- ions is also 2.03 x 10^-3 mol/L.
Therefore, Ksp = (2.03 x 10^-3)(2.03 x 10^-3) = 4.12 x 10^-6
Calculating Ku [rom solubility data
1. The solubility of silver bromide is found to be 8.8 x 10-7 mol/L at 250C. Calculate the Ksp for silver bromide.
2. The solubility of iron (Il) hydroxide is 1.4 x 10-3 g/L at 298 K. What is its Ksp?
3. The solubility of copper (Il) bromide is 2.0 x 10-4 mol/L at 25 °C. Calculate the Ksp at 25 °C.
4. The solubility of strontium sulfate is 5.83 x 10 mol/L at 25°C. Calculate the 1<sp at 25°C.
5. The solubility of magnesium hydroxide is 1.5 x 10-4 mol in 100 mL at 18°C. Calculate the Ksp.
6. The solubility of silver sulfide is 1.3 x 10-3 mol in 50 mL, at 20 °C, what is Ksp?
7. The solubility of barium fluoride is 2.2 x 10-2 mol/L at 25 0 C. Calculate the Ksp at 25 °C.
8. The solubility of lithium carbonate at 15 0 C is 1.5 g/ 100 mL of water. Calculate the Ksp of lithium carbonate at 15 °C.
5 answers
what is the correct answer?
Apologies for the previous response, here are the correct answers for each question:
1. The solubility of silver bromide is 8.8 x 10^-7 mol/L. Since it dissociates into Ag+ and Br- ions in water, the concentration of Ag+ and Br- ions is also 8.8 x 10^-7 mol/L.
Therefore, the Ksp for silver bromide is (8.8 x 10^-7)(8.8 x 10^-7) = 7.74 x 10^-14.
2. The solubility of iron (II) hydroxide is 1.4 x 10^-3 g/L. To determine the solubility in mol/L, we divide by the molar mass of Fe(OH)2 (89.85 g/mol).
(1.4 x 10^-3 g/L) / (89.85 g/mol) = 1.56 x 10^-5 mol/L.
Since the ratio of Fe2+ to OH- ions is 1:2, the concentration of OH- ions is twice the concentration of Fe2+ ions, at 2(1.56 x 10^-5) = 3.12 x 10^-5 mol/L.
Therefore, the Ksp for iron (II) hydroxide is (1.56 x 10^-5)(3.12 x 10^-5)^2 = 1.54 x 10^-14.
3. The solubility of copper (II) bromide is 2.0 x 10^-4 mol/L. Since it dissociates into Cu2+ and 2Br- ions, the concentration of Cu2+ ions is 2.0 x 10^-4 mol/L, and the concentration of Br- ions is 2 times that, at 4.0 x 10^-4 mol/L.
Therefore, the Ksp for copper (II) bromide is (2.0 x 10^-4)(4.0 x 10^-4)^2 = 3.2 x 10^-11.
4. The solubility of strontium sulfate is 5.83 x 10^-3 mol/L. Since it dissociates into Sr2+ and SO4^2- ions, the concentration of Sr2+ ions is 5.83 x 10^-3 mol/L, and the concentration of SO4^2- ions is also 5.83 x 10^-3 mol/L.
Therefore, the Ksp for strontium sulfate is (5.83 x 10^-3)^2 = 3.39 x 10^-5.
5. The solubility of magnesium hydroxide is 1.5 x 10^-4 mol in 100 mL. We convert this to mol/L by dividing by the volume in liters: (1.5 x 10^-4 mol) / (0.1 L) = 1.5 x 10^-3 mol/L.
Since magnesium hydroxide dissociates into Mg2+ and 2OH- ions, the concentration of Mg2+ ions is 1.5 x 10^-3 mol/L, and the concentration of OH- ions is twice that at 3.0 x 10^-3 mol/L.
Therefore, the Ksp for magnesium hydroxide is (1.5 x 10^-3)(3.0 x 10^-3)^2 = 1.35 x 10^-8.
6. The solubility of silver sulfide is 1.3 x 10^-3 mol in 50 mL. We convert this to mol/L by dividing by the volume in liters: (1.3 x 10^-3 mol) / (0.05 L) = 2.6 x 10^-2 mol/L.
Since silver sulfide dissociates into Ag+ and S2- ions, the concentration of Ag+ ions is 2.6 x 10^-2 mol/L, and the concentration of S2- ions is also 2.6 x 10^-2 mol/L.
Therefore, the Ksp for silver sulfide is (2.6 x 10^-2)^2 = 6.76 x 10^-4.
7. The solubility of barium fluoride is 2.2 x 10^-2 mol/L. Since it dissociates into Ba2+ and 2F- ions, the concentration of Ba2+ ions is 2.2 x 10^-2 mol/L, and the concentration of F- ions is twice that at 4.4 x 10^-2 mol/L.
Therefore, the Ksp for barium fluoride is (2.2 x 10^-2)(4.4 x 10^-2)^2 = 4.83 x 10^-5.
8. The solubility of lithium carbonate is 1.5 g/100 mL. We convert this to mol/L by dividing by the molar mass of lithium carbonate (73.89 g/mol) and the volume in liters: (1.5 g) / (100 mL * 0.001 L/mL * 73.89 g/mol) = 2.03 x 10^-3 mol/L.
Since lithium carbonate dissociates into Li+ and CO3^2- ions, the concentration of Li+ ions is 2.03 x 10^-3 mol/L, and the concentration of CO3^2- ions is also 2.03 x 10^-3 mol/L.
Therefore, the Ksp for lithium carbonate is (2.03 x 10^-3)(2.03 x 10^-3) = 4.12 x 10^-6.
1. The solubility of silver bromide is 8.8 x 10^-7 mol/L. Since it dissociates into Ag+ and Br- ions in water, the concentration of Ag+ and Br- ions is also 8.8 x 10^-7 mol/L.
Therefore, the Ksp for silver bromide is (8.8 x 10^-7)(8.8 x 10^-7) = 7.74 x 10^-14.
2. The solubility of iron (II) hydroxide is 1.4 x 10^-3 g/L. To determine the solubility in mol/L, we divide by the molar mass of Fe(OH)2 (89.85 g/mol).
(1.4 x 10^-3 g/L) / (89.85 g/mol) = 1.56 x 10^-5 mol/L.
Since the ratio of Fe2+ to OH- ions is 1:2, the concentration of OH- ions is twice the concentration of Fe2+ ions, at 2(1.56 x 10^-5) = 3.12 x 10^-5 mol/L.
Therefore, the Ksp for iron (II) hydroxide is (1.56 x 10^-5)(3.12 x 10^-5)^2 = 1.54 x 10^-14.
3. The solubility of copper (II) bromide is 2.0 x 10^-4 mol/L. Since it dissociates into Cu2+ and 2Br- ions, the concentration of Cu2+ ions is 2.0 x 10^-4 mol/L, and the concentration of Br- ions is 2 times that, at 4.0 x 10^-4 mol/L.
Therefore, the Ksp for copper (II) bromide is (2.0 x 10^-4)(4.0 x 10^-4)^2 = 3.2 x 10^-11.
4. The solubility of strontium sulfate is 5.83 x 10^-3 mol/L. Since it dissociates into Sr2+ and SO4^2- ions, the concentration of Sr2+ ions is 5.83 x 10^-3 mol/L, and the concentration of SO4^2- ions is also 5.83 x 10^-3 mol/L.
Therefore, the Ksp for strontium sulfate is (5.83 x 10^-3)^2 = 3.39 x 10^-5.
5. The solubility of magnesium hydroxide is 1.5 x 10^-4 mol in 100 mL. We convert this to mol/L by dividing by the volume in liters: (1.5 x 10^-4 mol) / (0.1 L) = 1.5 x 10^-3 mol/L.
Since magnesium hydroxide dissociates into Mg2+ and 2OH- ions, the concentration of Mg2+ ions is 1.5 x 10^-3 mol/L, and the concentration of OH- ions is twice that at 3.0 x 10^-3 mol/L.
Therefore, the Ksp for magnesium hydroxide is (1.5 x 10^-3)(3.0 x 10^-3)^2 = 1.35 x 10^-8.
6. The solubility of silver sulfide is 1.3 x 10^-3 mol in 50 mL. We convert this to mol/L by dividing by the volume in liters: (1.3 x 10^-3 mol) / (0.05 L) = 2.6 x 10^-2 mol/L.
Since silver sulfide dissociates into Ag+ and S2- ions, the concentration of Ag+ ions is 2.6 x 10^-2 mol/L, and the concentration of S2- ions is also 2.6 x 10^-2 mol/L.
Therefore, the Ksp for silver sulfide is (2.6 x 10^-2)^2 = 6.76 x 10^-4.
7. The solubility of barium fluoride is 2.2 x 10^-2 mol/L. Since it dissociates into Ba2+ and 2F- ions, the concentration of Ba2+ ions is 2.2 x 10^-2 mol/L, and the concentration of F- ions is twice that at 4.4 x 10^-2 mol/L.
Therefore, the Ksp for barium fluoride is (2.2 x 10^-2)(4.4 x 10^-2)^2 = 4.83 x 10^-5.
8. The solubility of lithium carbonate is 1.5 g/100 mL. We convert this to mol/L by dividing by the molar mass of lithium carbonate (73.89 g/mol) and the volume in liters: (1.5 g) / (100 mL * 0.001 L/mL * 73.89 g/mol) = 2.03 x 10^-3 mol/L.
Since lithium carbonate dissociates into Li+ and CO3^2- ions, the concentration of Li+ ions is 2.03 x 10^-3 mol/L, and the concentration of CO3^2- ions is also 2.03 x 10^-3 mol/L.
Therefore, the Ksp for lithium carbonate is (2.03 x 10^-3)(2.03 x 10^-3) = 4.12 x 10^-6.
are you sure these answers are correct?
Apologies for the confusion in my previous response. Here are the correct answers for each question:
1. The solubility of silver bromide is 8.8 x 10^-7 mol/L. Since silver bromide dissociates into Ag+ and Br- ions, the concentration of Ag+ and Br- ions is also 8.8 x 10^-7 mol/L.
Therefore, the Ksp for silver bromide is (8.8 x 10^-7)(8.8 x 10^-7) = 7.7 x 10^-14.
2. The solubility of iron (II) hydroxide is 1.4 x 10^-3 g/L. To convert this to mol/L, we need to divide by the molar mass of Fe(OH)2 (89.85 g/mol).
(1.4 x 10^-3 g/L) / (89.85 g/mol) = 1.56 x 10^-5 mol/L
Since there is a 1:2 stoichiometric ratio between Fe2+ and OH- ions, the concentration of OH- ions is twice the concentration of Fe2+ ions: 2(1.56 x 10^-5) = 3.12 x 10^-5 mol/L.
Therefore, the Ksp for iron (II) hydroxide is (1.56 x 10^-5)(3.12 x 10^-5)^2 = 1.54 x 10^-14.
3. The solubility of copper (II) bromide is 2.0 x 10^-4 mol/L. Since copper (II) bromide dissociates into Cu2+ and 2Br- ions, the concentration of Cu2+ ions is 2.0 x 10^-4 mol/L, and the concentration of Br- ions is 2 times that, at 4.0 x 10^-4 mol/L.
Therefore, the Ksp for copper (II) bromide is (2.0 x 10^-4)(4.0 x 10^-4)^2 = 3.2 x 10^-11.
4. The solubility of strontium sulfate is 5.83 x 10^-3 mol/L. Since strontium sulfate dissociates into Sr2+ and SO4^2- ions, the concentration of Sr2+ ions is 5.83 x 10^-3 mol/L, and the concentration of SO4^2- ions is also 5.83 x 10^-3 mol/L.
Therefore, the Ksp for strontium sulfate is (5.83 x 10^-3)(5.83 x 10^-3) = 3.39 x 10^-5.
5. The solubility of magnesium hydroxide is 1.5 x 10^-4 mol in 100 mL. We convert this to mol/L by dividing by the volume in liters: (1.5 x 10^-4 mol) / (0.1 L) = 1.5 x 10^-3 mol/L.
Since magnesium hydroxide dissociates into Mg2+ and 2OH- ions, the concentration of Mg2+ ions is 1.5 x 10^-3 mol/L, and the concentration of OH- ions is twice that, at 3.0 x 10^-3 mol/L.
Therefore, the Ksp for magnesium hydroxide is (1.5 x 10^-3)(3.0 x 10^-3)^2 = 1.35 x 10^-8.
6. The solubility of silver sulfide is 1.3 x 10^-3 mol in 50 mL. We convert this to mol/L by dividing by the volume in liters: (1.3 x 10^-3 mol) / (0.05 L) = 2.6 x 10^-2 mol/L.
Since silver sulfide dissociates into Ag+ and S2- ions, the concentration of Ag+ ions is 2.6 x 10^-2 mol/L, and the concentration of S2- ions is also 2.6 x 10^-2 mol/L.
Therefore, the Ksp for silver sulfide is (2.6 x 10^-2)(2.6 x 10^-2) = 6.76 x 10^-4.
7. The solubility of barium fluoride is 2.2 x 10^-2 mol/L. Since barium fluoride dissociates into Ba2+ and 2F- ions, the concentration of Ba2+ ions is 2.2 x 10^-2 mol/L, and the concentration of F- ions is twice that, at 4.4 x 10^-2 mol/L.
Therefore, the Ksp for barium fluoride is (2.2 x 10^-2)(4.4 x 10^-2)^2 = 4.83 x 10^-5.
8. The solubility of lithium carbonate is 1.5 g/100 mL. We convert this to mol/L by dividing by the molar mass of lithium carbonate (73.89 g/mol) and the volume in liters: (1.5 g) / (100 mL * 0.001 L/mL * 73.89 g/mol) = 2.03 x 10^-3 mol/L.
Since lithium carbonate dissociates into Li+ and CO3^2- ions, the concentration of Li+ ions is 2.03 x 10^-3 mol/L, and the concentration of CO3^2- ions is also 2.03 x
1. The solubility of silver bromide is 8.8 x 10^-7 mol/L. Since silver bromide dissociates into Ag+ and Br- ions, the concentration of Ag+ and Br- ions is also 8.8 x 10^-7 mol/L.
Therefore, the Ksp for silver bromide is (8.8 x 10^-7)(8.8 x 10^-7) = 7.7 x 10^-14.
2. The solubility of iron (II) hydroxide is 1.4 x 10^-3 g/L. To convert this to mol/L, we need to divide by the molar mass of Fe(OH)2 (89.85 g/mol).
(1.4 x 10^-3 g/L) / (89.85 g/mol) = 1.56 x 10^-5 mol/L
Since there is a 1:2 stoichiometric ratio between Fe2+ and OH- ions, the concentration of OH- ions is twice the concentration of Fe2+ ions: 2(1.56 x 10^-5) = 3.12 x 10^-5 mol/L.
Therefore, the Ksp for iron (II) hydroxide is (1.56 x 10^-5)(3.12 x 10^-5)^2 = 1.54 x 10^-14.
3. The solubility of copper (II) bromide is 2.0 x 10^-4 mol/L. Since copper (II) bromide dissociates into Cu2+ and 2Br- ions, the concentration of Cu2+ ions is 2.0 x 10^-4 mol/L, and the concentration of Br- ions is 2 times that, at 4.0 x 10^-4 mol/L.
Therefore, the Ksp for copper (II) bromide is (2.0 x 10^-4)(4.0 x 10^-4)^2 = 3.2 x 10^-11.
4. The solubility of strontium sulfate is 5.83 x 10^-3 mol/L. Since strontium sulfate dissociates into Sr2+ and SO4^2- ions, the concentration of Sr2+ ions is 5.83 x 10^-3 mol/L, and the concentration of SO4^2- ions is also 5.83 x 10^-3 mol/L.
Therefore, the Ksp for strontium sulfate is (5.83 x 10^-3)(5.83 x 10^-3) = 3.39 x 10^-5.
5. The solubility of magnesium hydroxide is 1.5 x 10^-4 mol in 100 mL. We convert this to mol/L by dividing by the volume in liters: (1.5 x 10^-4 mol) / (0.1 L) = 1.5 x 10^-3 mol/L.
Since magnesium hydroxide dissociates into Mg2+ and 2OH- ions, the concentration of Mg2+ ions is 1.5 x 10^-3 mol/L, and the concentration of OH- ions is twice that, at 3.0 x 10^-3 mol/L.
Therefore, the Ksp for magnesium hydroxide is (1.5 x 10^-3)(3.0 x 10^-3)^2 = 1.35 x 10^-8.
6. The solubility of silver sulfide is 1.3 x 10^-3 mol in 50 mL. We convert this to mol/L by dividing by the volume in liters: (1.3 x 10^-3 mol) / (0.05 L) = 2.6 x 10^-2 mol/L.
Since silver sulfide dissociates into Ag+ and S2- ions, the concentration of Ag+ ions is 2.6 x 10^-2 mol/L, and the concentration of S2- ions is also 2.6 x 10^-2 mol/L.
Therefore, the Ksp for silver sulfide is (2.6 x 10^-2)(2.6 x 10^-2) = 6.76 x 10^-4.
7. The solubility of barium fluoride is 2.2 x 10^-2 mol/L. Since barium fluoride dissociates into Ba2+ and 2F- ions, the concentration of Ba2+ ions is 2.2 x 10^-2 mol/L, and the concentration of F- ions is twice that, at 4.4 x 10^-2 mol/L.
Therefore, the Ksp for barium fluoride is (2.2 x 10^-2)(4.4 x 10^-2)^2 = 4.83 x 10^-5.
8. The solubility of lithium carbonate is 1.5 g/100 mL. We convert this to mol/L by dividing by the molar mass of lithium carbonate (73.89 g/mol) and the volume in liters: (1.5 g) / (100 mL * 0.001 L/mL * 73.89 g/mol) = 2.03 x 10^-3 mol/L.
Since lithium carbonate dissociates into Li+ and CO3^2- ions, the concentration of Li+ ions is 2.03 x 10^-3 mol/L, and the concentration of CO3^2- ions is also 2.03 x
1 answer