What is the silver ion concentration in a solution prepared by mixing 345 mL of 0.352 M silver nitrate with 443 mL of 0.511 M sodium carbonate? The Ksp of silver carbonate is 8.1 × 10-12
This is a limiting reagent problem, a common ion problem, and a Ksp problem all rolled into one.
345 mL x 0.352M AgNO3 = 121.44 mmoles. We'll round this to 121 mmols AgNO3.
443 mL x 0.511M Na2CO3 = 226.373 mmols. We'll round this to 226 mmols Na2CO3.
You can round them differently and/or carry more places if you wish.
2AgNO3 + Na2CO3 ==> Ag2CO3 + 2NaNO3
First we find the limiting reagent
Convert 121 mmols AgNO3 to mmols Ag2CO3. That's 121 x (mols Ag2CO3/2 mols AgNO3) = 121 x (1/2) = 60.5 mmols Ag2CO3.
Convert 226 mmols Na2CO3 to Ag2CO3. That's 226 x (1 mol Ag2CO3/1 mol Na2CO3) = 226 x 1/1 = 226.
Obviously both answers can be right; the correct answer in limiting reagent problems is ALWAYS the smaller one and the reagent producing that value is the limiting reagent. Therefore, AgNO3 is the limiting reagent.
Next we determine how much of the Na2CO3 is used.
mmols Na2CO3 used = 121 mmols AgNO3 x (1 mol Na2CO3/2 mol AgNO3) = 121 x (1/2) = 60.5 mmols used. So how much is left unused? That's 226 - 60.5 = 165.5.
That 165.5 mmols Na2CO3 is the common ion. What's the concn now? We have 165.5 mmols and it's in a volume of 345 mL + 443 mL = 788 mL so the concn is
M = mmols/mL = 165.5/788 = about 0.2 but you do it more accurately than that.
.........Ag2CO3 ==> 2Ag^+ + CO3^2-
I.........solid......0......0.2
C.........solid.....+2x.....+x
E.........solid.....2x....0.2+x
Ksp = (Ag^+)^2(CO3^2-)
8.1E-12 = (2x)^2(0.2+x)
Solve for x, then 2x = (Ag^+).
These problems are too long. I wonder if this is for a general chemistry class or an advanced class.
To find the silver ion concentration in the mixed solution, we need to determine if a precipitation reaction will occur between silver nitrate and sodium carbonate.
Step 1: Write the balanced equation for the reaction:
2AgNO3 + Na2CO3 → Ag2CO3 + 2NaNO3
In this reaction, silver nitrate (AgNO3) reacts with sodium carbonate (Na2CO3) to form silver carbonate (Ag2CO3) and sodium nitrate (NaNO3).
Step 2: Calculate the number of moles of silver nitrate:
Moles of AgNO3 = Volume of silver nitrate × Concentration of silver nitrate
= 345 mL × 0.352 M
= 121.44 mmol (millimoles)
Step 3: Calculate the number of moles of sodium carbonate:
Moles of Na2CO3 = Volume of sodium carbonate × Concentration of sodium carbonate
= 443 mL × 0.511 M
= 226.33 mmol
Step 4: Calculate the limiting reactant:
To determine which reactant is the limiting reactant, we need to compare the moles of each reactant and see which one produces the least amount of product. From the balanced equation, we can see that 2 moles of AgNO3 react with 1 mole of Na2CO3 to produce 1 mole of Ag2CO3. Therefore, the ratio of AgNO3 to Na2CO3 is 2:1.
Moles of AgNO3 in excess = 2 × Moles of Na2CO3
= 2 × 226.33 mmol
= 452.66 mmol
Since we only have 121.44 mmol of AgNO3, it is the limiting reactant.
Step 5: Calculate the moles of silver carbonate formed:
From the balanced equation, we can see that 2 moles of AgNO3 react to produce 1 mole of Ag2CO3. Therefore, the number of moles of Ag2CO3 formed is equal to the number of moles of AgNO3 used.
Moles of Ag2CO3 formed = Moles of AgNO3 used
= 121.44 mmol
Step 6: Calculate the concentration of silver ions:
The concentration of silver ions in the solution is equal to the moles of Ag2CO3 formed divided by the total volume of the solution.
Total volume of the solution = Volume of silver nitrate + Volume of sodium carbonate
= 345 mL + 443 mL
= 788 mL
Concentration of silver ions = Moles of Ag2CO3 formed / Total volume of the solution
= (121.44 mmol) / (788 mL)
= 0.154 M
Therefore, the silver ion concentration in the solution is 0.154 M.