12. At 50°C, the solubility of silver chloride, AgCl, is 5.2 x 10g/100 mL. Calcu
ID: 556215 • Letter: 1
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12. At 50°C, the solubility of silver chloride, AgCl, is 5.2 x 10g/100 mL. Calculate the Kue at this temperature. Answer: 1.3 × 10-9 13. If 1.00 g of MgF2 (FW = 62.31) is added to 500, ml of pure water, how many grams dissolve? Kp7.4 x 10-9 at 19°C) -Answer: 0.038 14. The addition of fluoride to many public water systems is done to help fight cavities in a city's population. Given a 500 mL sample of hard water that contains 2.5 x 10-3M Ca2t ions, what is the maximum amount of sodium fluoride, in mg, that can be dissolved in the water with no precipitation? (Kg [CaF2-5.3 × 10- Answer. 30.6 mg 15. To a 20 L solution with b2. = 0,0010 M is added sufficient potassium sulfate to initiate precipitation. How much KS04 was added? The molar mass of potassium sulfate is 1742 g/mol. (For Pbso, , Ksp = 7.1 x 10-") Answer: 0.025 g 16. A saturated solution of AgCrOs contains 14 mg AgCrOs per liter of solution. What is Kip for AgCrO? Answer: 4.5 × 10-9 17. What is the molar solubility of Pbl in a solution that contains 0.30 M KIP (Kp 79 x 10) Answer: 8.8 × 10-8 18. ) If 100. ml. of 0.025 M NaCl is mixed with 150. mL of 0.0015 M Pb(NO, will a precipitate form? (K PbCl,-1.6 × 10-5) Answer No- Justify this answer) 1%) Calculate the concentration of free Ag+ in a solution containing 0.10 M Ag+ and 0.30 M SO (K( (Ag(SA)= 1.7 × 1013) Answer: 2.9×10-14M 20. A saturated solution of AB2 contains 8.60 x 10-4 moles AB2 per liter. Determine Kp 21. For this equilibrium, What would be the effect of making silver undergo a second reaction with ammonia? 22. The Ksp for ZnS is 1.0x10-1 at 25 C Answer: 2.54 x 10-9 Agl (s)Ag (aq) +I (aq) By what factor will the solubility of ZnS increase in 0.50 M solution of NHs compared to pure water? K, for Zn(NHs)?' = 7.8x10"? Answer: It will be close to 7000 times more soluble (Divide the solubility in NHs by the solubility in water)Explanation / Answer
According to chegg rules we have to answer only first one problem
12.
Given that the solubility of AgCl at 50 C is 5.2 x 10^-4 g/100 mL.
First convert g to moles as follows:
Number of moles = amount in g/ molar mass
= 5.2*10^-4 g/ 143.32 g/ mol
= 3.6*10^-6 mole /100 ml
Now calculate the moalrity as follows:
Molarity = number of moles / volume in L
= 3.6*10^-6 mole /100 ml *1.0 L/1000 ml
= 3.6*10^-5 mole / L or M
Solubility poridust is given as follows:
AgCl = Ag+ + Cl-,
Ksp = [Ag+] [Cl-]
= (3.6*10^-5) (3.6*10^-5)
= 1.3*10^-9
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