Question: the conversion of one precipitate to another by the addition of a suitable reagent is a process that involves heterogeneous equilibrium system. for example, silver chloride can be coverted to silver bromide by the additino of aqueous sodium bromide to a sample of silver chloride. the net conversion is AgCL(s) + Br goes <-> AgBr(s) +Cl(aq).
How many grams of AgCl will be converted to AgBr by using 100.9ml of aqueous .100 M. NaBr?
Would you set up an ice table? How do you use the numbers given? Do you plug the .100 M of NaBr and than solve using x for Cl?
Notice in the reaction, each mole of Br converts one mole of Cl (not diatomic).
so if x is the amountof Br in AgBr, then x is also the amount of Cl on the right.
Yes, use x for this.
I can't find a Ksp value for this problem. I found a complex ion kf value for it though. How do i compute the molar solubility assuming hydrolysis occurs.
oops never mind it doesn't work that way
Please check my answer:
once i set up the ICE table i get this: the ksp= 1.8e-10= x/(.100-X)
x= 1.8e-9 M
1.8e-9 M * .1000L = 1.8e-10 mol (143.3212 g AgCl/ 1 mol)= 2.58e-8g AgCl.
Is this correct?
To solve this question, you would need to set up an ICE table and use the given information. Here's how you can approach the problem:
1. Start by determining the balanced chemical equation for the conversion:
AgCl(s) + Br(aq) ⇌ AgBr(s) + Cl(aq)
2. Since we are given the volume and concentration of NaBr, we can use this information to find the moles of Br(aq) present in the solution. Remember that moles = concentration × volume (in liters). The given volume is 100.9 mL = 0.1009 L, and the concentration is 0.100 M. So, moles of Br(aq) = 0.100 M × 0.1009 L = 0.01009 moles.
3. Now, construct an ICE table. 'I' represents the initial moles or concentration, 'C' represents the change in moles or concentration, and 'E' represents the equilibrium moles or concentration. Since we are dealing with a solid (AgBr) and an aqueous species (Cl-), we can ignore the solid's concentration in the equilibrium expression.
Initial:
AgCl(s): X (assume 'X' mol)
Br(aq): 0.01009 moles
Change:
AgCl(s): -X (since X mol of AgCl is converted)
Br(aq): -X (stoichiometry of the balanced equation)
Equilibrium:
AgCl(s): X - X = 0 mol
Br(aq): 0.01009 - X moles
4. The equilibrium expression for this reaction is given as:
Kc = ([AgBr(s)] × [Cl(aq)]) / ([AgCl(s)] × [Br(aq)])
Since the concentration of AgCl is X and the concentration of Br(aq) is 0.01009 - X, the equilibrium expression becomes:
Kc = ([AgBr(s)] × [Cl(aq)]) / (X × (0.01009 - X))
5. At equilibrium, the concentrations remain constant. Since AgBr is a solid, its concentration remains constant and equals zero. Therefore, the equilibrium expression can be simplified to:
0 = X × (0.01009 - X)
6. Solve the equation to find the value of X. Rearrange the equation to:
0.01009X - X^2 = 0
This is a quadratic equation. Solve it by factoring or using the quadratic formula. By solving for X, you will find the moles of AgCl that are converted to AgBr.
7. Finally, calculate the mass of AgCl that will be converted. Convert the moles of AgCl to grams using the molar mass of AgCl, and that will give you the answer to the question.
Remember to check your calculations and units throughout the steps.