A different solution contains dissolved NaIO3 . What is the concentration of NaIO3 if adding excess Pb(IO3)2(s) produces a Pb2+ concentration of 5.40×10−6 M ?
I did that already yet.
Well, if I had a nickel for every time I tried to solve an equation with excessive amounts of chemistry involved, I'd have... well, about five cents. Which is still more than zero, so hey, I'll take it!
Now, let's get our clown noses on and go through this. We have NaIO3, which is all dissolved up, and we're adding excess Pb(IO3)2(s) to it. This reaction is going to be a bit like a drama queen, because it's a double replacement reaction. So, we can expect some fancy swapping of ions here.
First, let's balance the equation:
2NaIO3 + Pb(IO3)2 → 2Na(IO3)2 + PbI2
Now, my friend, pay attention to the excess part. It means that we have more than enough Pb(IO3)2 to react with all the NaIO3. So, the limiting reagent in this circus is going to be the NaIO3.
Since we know the concentration of Pb2+ is 5.40×10−6 M, we can use stoichiometry to find the concentration of NaIO3. From the balanced equation, we can see that 2 moles of NaIO3 give us 1 mole of Pb2+.
So, if the concentration of Pb2+ is 5.40×10−6 M, that means the concentration of NaIO3 is going to be half that, since we have a 2 to 1 ratio. So, the concentration of NaIO3 is 2.70×10−6 M.
I hope that answered your question, and that you enjoyed this chemistry circus act. If you need any more laughs, or have more questions, just drop me a line!
To determine the concentration of NaIO3, we can use the balanced chemical equation between NaIO3 and Pb(IO3)2:
2 NaIO3 + Pb(IO3)2 → 2 NaIO3 + PbI2
Here, we can see that two moles of NaIO3 react with one mole of Pb(IO3)2 to produce two moles of NaIO3. This means the mole ratio between NaIO3 and Pb(IO3)2 is 2:1.
Given that the Pb2+ concentration is 5.40×10−6 M, we can assume that all of the Pb(IO3)2 has dissolved to produce this concentration. Therefore, the concentration of Pb2+ is directly equal to the concentration of Pb(IO3)2.
Since the mole ratio between NaIO3 and Pb(IO3)2 is 2:1, the concentration of NaIO3 will be half the concentration of Pb(IO3)2.
Concentration of NaIO3 = 0.5 × 5.40×10−6 M = 2.70×10−6 M
So, the concentration of NaIO3 is 2.70×10−6 M.
To find the concentration of NaIO3, we need to use stoichiometry and the balanced chemical equation for the reaction between NaIO3 and Pb(IO3)2. Here's how you can approach this problem step by step:
Step 1: Write the balanced chemical equation for the reaction between NaIO3 and Pb(IO3)2:
5NaIO3 + Pb(IO3)2 → 3NaIO3 + 2Pb(IO3)2
Step 2: Determine the stoichiometric ratio between NaIO3 and Pb(IO3)2. From the balanced equation, we can see that 5 moles of NaIO3 react with 1 mole of Pb(IO3)2.
Step 3: Calculate the moles of Pb2+ ions produced. We are given that the concentration of Pb2+ is 5.40×10−6 M. Since the stoichiometric ratio between Pb(IO3)2 and Pb2+ ions is 1:2, the concentration of Pb(IO3)2 is also 5.40×10−6 M.
Step 4: Use the stoichiometric ratio to find the moles of NaIO3. Since the ratio is 5:1, the moles of NaIO3 are five times the moles of Pb(IO3)2.
Step 5: Convert the moles of NaIO3 to concentration. To do this, we need to know the volume of the solution. Without the volume information, we cannot calculate the concentration of NaIO3.
In conclusion, we have determined the stoichiometric ratio and found the moles of NaIO3 needed to react with the given concentration of Pb2+. However, we do not have the necessary information to calculate the concentration of NaIO3 without knowing the volume of the solution.