When 50 mL of 1.0 M AgNO3 is added to 50 mL of 0.50 M HCl, a precipitate of AgCl forms. After the reaction is complete, what is the concentration of silver ions in the resulting solution?
(a) 0.50 M
(b) 1.0 M
(c) zero
(d) 0.25 M
(e) 0.75 M
How do I go about solving this?
I attempted to start it.. I'm looking to find concentration, correct? So I would use molarity, which = moles solute / liters solution
M= mol solute / L soln
How do I find the moles of solute? which is the solute, is it AgNO3?
Would the L of soln be 0.1 L, because 50 mL + 50 mL = 100 mL = 0.1 L?
Please help!
Yes, the volume will be 0.05L + 0.05L= 0.1 L.
If you started with 0.05 L of 1 M AgNO3, you had 0.05 moles Ag ion initially.
You added 0.05 L x 0.5 M = 0.025 moles HCl.
So 0.025 moles Ag ion were used leaving
0.050-0.025 = 0.025 moles Ag ion (and I'm assuming we neglect the VERY small amount of Ag ion that comes from the AgCl ppt).
M = moles/L. 0.025/0.1 =
Well, well, well! It seems like you've stumbled upon a chemistry problem, my friend. Don't worry, I'll do my best to help you out and bring some humor to the equation.
First, let's identify the solute, which in this case is indeed AgNO3. Good job! And yes, the total volume of the solution would be 0.1 L since you combined 50 mL of AgNO3 and 50 mL of HCl.
Now, the magic starts when AgNO3 and HCl throw a little party and form AgCl, a lovely precipitate. So, you need to figure out how many moles of silver ions are present in that reaction mixture after the reaction is complete.
To do that, you'll need to use stoichiometry, my dear problem-solver. The balanced equation for the reaction is:
AgNO3 + HCl → AgCl + HNO3
Oh, the wonders of chemical equations! Now, let’s compare the coefficients of AgNO3 and AgCl. You'll notice that the ratio is 1:1 - for every mole of AgNO3, you get one mole of AgCl.
Since you've got 50 mL of 1.0 M AgNO3, you can use the equation M = mol solute / L soln to find the moles of AgNO3. So, M = 1.0 mol/L. Considering you've got 0.1 L of solution, just multiply those numbers together. Go ahead, I'll wait.
Once you've got the number of moles of AgNO3, you can confidently say that you have the same number of moles of Ag+ ions because of that lovely 1:1 ratio. So, the concentration of silver ions in the resulting solution is... drumroll, please...
(c) zero!
Yes, the Clowns of Chemistry have played their trick. You see, AgCl is insoluble in water, so it precipitates out of the solution, leaving no silver ions floating around. Sneaky, right?
I hope my silly explanation helps you understand the problem. Keep up the good work, my friend, and keep those chemistry questions coming!
To solve this question, let's first write out the balanced chemical equation for the reaction between AgNO3 (silver nitrate) and HCl (hydrochloric acid):
AgNO3 + HCl -> AgCl + HNO3
From the balanced equation, we can see that 1 mole of AgNO3 reacts with 1 mole of HCl to form 1 mole of AgCl.
Next, calculate the moles of AgNO3 and HCl separately using the given concentrations and volumes:
Moles of AgNO3 = Concentration of AgNO3 × Volume of AgNO3 solution
= 1.0 M × 0.05 L
= 0.05 moles
Moles of HCl = Concentration of HCl × Volume of HCl solution
= 0.50 M × 0.05 L
= 0.025 moles
Since AgNO3 and HCl react in a 1:1 ratio, the limiting reactant is HCl because it has fewer moles. This means that all the HCl will react, and AgNO3 will be in excess.
When HCl reacts with AgNO3, it forms AgCl. Therefore, the moles of AgCl formed will be equal to the moles of HCl used.
Moles of AgCl = Moles of HCl
= 0.025 moles
To find the concentration of silver ions in the resulting solution, we need to determine the final volume of the solution after the reaction. Since 50 mL of AgNO3 and 50 mL of HCl were mixed, the final volume will be 100 mL or 0.1 L.
Finally, we can calculate the concentration of silver ions:
Concentration of Ag ions = Moles of AgCl / Volume of resulting solution
= 0.025 moles / 0.1 L
= 0.25 M
Therefore, the concentration of silver ions in the resulting solution is 0.25 M.
The correct answer is (d) 0.25 M.
To solve this problem, you need to calculate the concentration of silver ions in the resulting solution.
First, let's identify the balanced equation for the reaction between silver nitrate (AgNO3) and hydrochloric acid (HCl):
AgNO3 + HCl → AgCl + HNO3
From the balanced equation, you can see that silver nitrate reacts with hydrochloric acid to produce silver chloride (AgCl) and nitric acid (HNO3).
Since we know the initial concentrations and volumes of both solutions, we can use the concept of stoichiometry to determine the concentration of silver ions in the resulting solution.
Step 1: Calculate the number of moles of AgNO3 used.
Molarity (M) = moles of solute / liters of solution
Given that 50 mL of 1.0 M AgNO3 is used, convert the volume to liters:
50 mL = 50 mL * (1 L / 1000 mL) = 0.05 L
Now calculate the number of moles of AgNO3:
moles of AgNO3 = Molarity * Volume
moles of AgNO3 = 1.0 M * 0.05 L = 0.05 moles
Step 2: Calculate the number of moles of HCl used.
The stoichiometry of the balanced equation tells us that AgNO3 and HCl react in a 1:1 ratio. Therefore, the number of moles of HCl used is also 0.05 moles.
Step 3: Determine the limiting reactant.
Since AgNO3 and HCl are in equal amounts (0.05 moles each), neither of them is in excess. The limiting reactant is the one that produces the smallest amount of product. In this case, AgNO3 and HCl react in a 1:1 ratio to produce AgCl. So, the limiting reactant is either AgNO3 or HCl.
Step 4: Calculate the number of moles of AgCl formed.
Since the limiting reactant is in a 1:1 ratio with AgCl, the number of moles of AgCl formed is equal to the number of moles of the limiting reactant, which is 0.05 moles.
Step 5: Calculate the concentration of silver ions in the resulting solution.
The resulting solution contains both the excess reagent (the one that was not fully consumed) and the product, AgCl.
The excess HCl will remain in solution, and its concentration will be the same as the initial concentration, which is 0.50 M.
Since AgCl is a precipitate, it is no longer in the solution, so its concentration is zero.
Therefore, the concentration of silver ions in the resulting solution is also zero (option c).
To summarize:
After the reaction is complete, the concentration of silver ions in the resulting solution is zero (option c).