If 100.0 mL of 0.250 M KMnO4 is mixed with 300.0 mL of 0.500 M KMnO4, what is the concentration of the final..

Question

mixture?

I have the answer (0.438) but I can't find how to get it. Equation and steps plx.

Answer 1

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Answer 2

Moles in first solution is

100 ML/1000(mL/L)x.250mol/L = .025 mol

Moles in second solution is:

300 mL/1000(mL/L)x.500 mol/L= .150 mol

Add number of moles and divide to the total volume of solution:

(0.025 mol + 0.150 mol) / (.1L + .3L)=0.438 mol/L