you add 0.05ml of 0.10M agno3 to 4ml of 2.0 nacl it forms saturated solution of agcl<---> ag+ + cl-
using a saturated solution you set up
ag | ag+( test solution after reaction) || ag+(1.0ml) | ag
how do we calc the equilibrium of ag+ of the test solution after reaction ?
Ecell = Eocell -(0.0592/2)log(x/1) and solve for x.
is the Eo cell =0 in this case?
yes
Shouldn't the n be 1?
To calculate the equilibrium concentration of Ag+ in the test solution after the reaction, you need to know the initial concentrations of Ag+ and Cl- and the equilibrium constant (K) for the reaction.
1. Initial concentrations:
- Ag+ initial concentration: 0.05 mL of 0.10 M AgNO3 = 0.01 mmol of Ag+
- Cl- initial concentration: 4 mL of 2.0 M NaCl = 8 mmol of Cl-
2. Write the balanced chemical equation:
AgNO3(aq) + NaCl(aq) ⇌ AgCl(s) + NaNO3(aq)
The solubility product constant (Ksp) expression for this reaction is:
Ksp = [Ag+][Cl-]
3. Calculate the concentration of Ag+ at equilibrium:
Let's assume x is the change in Ag+ concentration. The final concentration of Ag+ will be (0.01 - x) mmol.
The concentration of Cl- at equilibrium will be (8 - x) mmol.
Therefore, the equilibrium expression becomes:
Ksp = (0.01 - x)(8 - x)
4. Solve for x:
Rearrange the equation and substitute Ksp (which you can find in a reference source) to solve for x:
(0.01 - x)(8 - x) = Ksp
Solve the quadratic equation to find the value of x, which represents the change in Ag+ concentration.
5. Calculate the equilibrium concentration of Ag+:
The equilibrium concentration of Ag+ is given by (0.01 - x) mmol.
Note: The actual numerical calculation will depend on the value of the solubility product constant (Ksp) specific to AgCl.