You have 3.00 L of a 3.00 M solution of NaCl(aq) called solution A. You also have 2.00 L of a 2.00 M solution of AgNO3(aq) called solution B. You mix these solutions together, making solution C. Hint: AgCl is a precipitate.
Calculate the concentrations (in M) of the following ions in solution C.
Cl-
Group of answer choices
0 M
1.00 M
0.800 M
1.80 M
1.50 M
To determine the concentration of Cl- in solution C, we need to consider the stoichiometry of the reaction between NaCl and AgNO3.
The balanced chemical equation for the reaction is:
NaCl(aq) + AgNO3(aq) -> AgCl(s) + NaNO3(aq)
From the equation, we can see that 1 mole of AgCl is produced for every mole of Cl- in the reactants.
First, let's calculate the moles of Cl- in solution A.
moles of Cl- in A = (volume of solution A) x (concentration of Cl- in A)
= (3.00 L) x (3.00 M)
= 9.00 moles
Since the stoichiometry of the reaction is 1:1 between Cl- and AgCl, we know that 9.00 moles of Cl- will react with 9.00 moles of AgNO3.
Next, let's calculate the moles of AgNO3 in solution B.
moles of AgNO3 in B = (volume of solution B) x (concentration of AgNO3 in B)
= (2.00 L) x (2.00 M)
= 4.00 moles
Since 1 mole of AgNO3 produces 1 mole of AgCl, we know that 4.00 moles of AgNO3 will produce 4.00 moles of AgCl.
However, since AgCl is insoluble and forms a precipitate, it will be removed from the solution and not contribute to the final concentration of Cl-.
Therefore, the remaining moles of Cl- in solution C is:
moles of Cl- in C = moles of Cl- in A - moles of AgNO3 in B
= 9.00 moles - 4.00 moles
= 5.00 moles
Finally, we calculate the concentration of Cl- in solution C by dividing the moles of Cl- by the total volume of solution C.
concentration of Cl- in C = (moles of Cl- in C) / (total volume of solution C)
= 5.00 moles / (3.00 L + 2.00 L)
= 5.00 moles / 5.00 L
= 1.00 M
Therefore, the concentration of Cl- in solution C is 1.00 M.
The correct option is: 1.00 M