How much of a 1.25 M sodium chloride solution in milliliters is required to completely precipitate all of the silver in 25.0 mL of a 0.45 M silver nitrate solution?
Problem 117P :
Molarity of Sodium chloride(NaCl) solution = 1.25 M
Molarity of silver nitrate(AgNO3) solution = 0.45 M
Volume of silver nitrate(AgNO3) solution = 25.0 mL
Volume of solution is in mL, so let’s convert to L :
1 L = 1000 mL
Therefore, 25.0 mL in L will be :
= 25 mL
= 0.025 L.
First, Let’s write the chemical reaction that’s happening :
NaCl (aq) + AgNO3 (aq) → NaNO3 (aq) + AgCl (s)
Now, let’s find the number of moles of AgNO3 in solution :
Here, mass of AgNO3 is not given. Hence, we shall calculate the number of moles using the molarity formula as AgNO3 is the solute :
Therefore, the number of moles(m) can be derived from the above formula as :
Moles of solute = Molarity Volume of Solution (in L)
Therefore , moles of AgNO3 = 0.45 M0.025 L
= 0.01125 moles
Now, let’s Find the number of moles of sodium chloride(NaCl):
In the reaction, we see that the ratio of solute and solvent is 1:1. Hence, there will be
0.01125 moles of sodium chloride(NaCl).