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.
Ag+
Group of answer choices
0 M
1.80 M
0.800 M
1.50 M
1.00 M
To find the concentration of Ag+ in solution C, we need to consider the stoichiometry of the reaction between NaCl and AgNO3. When these two solutions are mixed together, a double displacement reaction occurs, leading to the formation of AgCl precipitate and NaNO3:
NaCl(aq) + AgNO3(aq) -> AgCl(s) + NaNO3(aq)
From the balanced equation, we can see that one mole of AgNO3 reacts with one mole of NaCl to produce one mole of AgCl.
First, let's calculate the number of moles of NaCl and AgNO3 present in the initial solutions:
Moles of NaCl in solution A = volume × concentration = 3.00 L × 3.00 M = 9.00 mol
Moles of AgNO3 in solution B = volume × concentration = 2.00 L × 2.00 M = 4.00 mol
Since we have equal moles of NaCl and AgNO3, all of the NaCl will react with AgNO3, resulting in the complete precipitation of Ag+ ions as AgCl.
The volume of the final solution C is the sum of the volumes of the initial solutions A and B:
Volume of solution C = volume of solution A + volume of solution B = 3.00 L + 2.00 L = 5.00 L
The number of moles of Ag+ ions in solution C can be found by dividing the number of moles of AgNO3 by the volume of solution C:
Moles of Ag+ in solution C = 4.00 mol / 5.00 L = 0.80 M
Therefore, the concentration of Ag+ ions in solution C is 0.80 M. Thus, the correct answer is 0.800 M.