Respuesta :
Answer:
Since Q > Ksp, a precipitate of AgCl will form.
Explanation:
Step 1: Data given
Molarity AgNO3 = 0.10 M
Molarity of NaCl = 0.075 M
Ksp AgCl = 1.77 * 10^-10
Step 2: The balanced equation
AgNO3 + NaCl → AgCl(s) + NaNO3(aq)
For 0.10 moles AgNO3 we have 0.10 moles Ag+ ( molarity = 0.10 M)
For 0.075 moles NaCl we have 0.075 moles Cl- (molarity = 0.075 M)
Step 3: Calculate Q
Q = [Ag+][Cl-]
Q = (0.10 M )(0.075 M ) = 0.0075
Ksp = 1.77 * 10^-10
Q >>> Ksp
Since Q > Ksp, a precipitate of AgCl will form.
When The mixed with a solution of NaCl to form 0.10 M in AgNO3 and 0.075 M in NaCl. Ksp (AgCl) is = 1.77 × 10-10 Since Q > Ksp,a precipitate of AgCl will form.
Computation of Molarity AgNO3
Data given as per question:
Then Molarity AgNO3 is = 0.10 M
After that Molarity of NaCl is = 0.075 M
Then Ksp AgCl is = 1.77 * 10^-10
Then The balanced equation are:
Formula is AgNO3 + NaCl → AgCl(s) + NaNO3(aq)
Then For 0.10 moles AgNO3 we have a 0.10 moles Ag+ ( molarity is = 0.10 M)
After that For 0.075 moles NaCl we have a 0.075 moles Cl- (molarity is = 0.075 M)
Now we Calculate Q that is:
Then Q is = [Ag+][Cl-]
After that Q is = (0.10 M )(0.075 M ) = 0.0075
Now, Ksp is = 1.77 * 10^-10
Therefore, Q >>> Ksp
Find more information about Molarity AgNO3 here:
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