What formula would you use to solve this problem? I'm stuck.
You have prepared a solution by diluting 12.00mL of 0.790M copper(II) nitrate to a total volume of 50.00mL. What is the concentration of Cu2+ ions in the resulting solution? Report your answer to 3 decimal places.
mL1 x M2 = mL2 x M2
12 x 0.790 = 50 x M2
OR, since it's dilution, you start with 0.790M and multiply by a factor less than 1 like this
0.790 x 12/50 = ?
Technically, M2 is the concentration of Cu(NO3)2 but since 1 mol Cu = 1 mol Cu(NO3)2 it's all the same.
You have found that 14.37mL of 0.162M sulfuric acid reacts with exactly 20.00mL of barium hydroxide solution of unknown concentration. What is the molarity of the barium hydroxide solution? Round your answer to 3 decimal places and include units.
I know that this has something to do with Molarity and that the molarity equation is (M) = moles of solute / volume of solution (in liters) What I'm unsure of is how do set it up wit the unknown?
To solve this problem, you need to use the formula for dilution:
C1V1 = C2V2
Where,
C1 = initial concentration of the solution
V1 = initial volume of the solution
C2 = final concentration of the solution
V2 = final volume of the solution
In this problem, we have:
C1 = 0.790 M (initial concentration of copper(II) nitrate)
V1 = 12.00 mL (initial volume of copper(II) nitrate)
C2 = ? (final concentration of Cu2+ ions)
V2 = 50.00 mL (final volume after dilution)
We need to solve for C2:
C1V1 = C2V2
0.790 M * 12.00 mL = C2 * 50.00 mL
Rearranging the equation to solve for C2:
C2 = (C1V1) / V2
Substituting the given values:
C2 = (0.790 M * 12.00 mL) / 50.00 mL
Calculating the answer:
C2 = 0.1896 M
Therefore, the concentration of Cu2+ ions in the resulting solution is 0.189 M (rounded to 3 decimal places).