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# Volume of K_2 SO_4 to be added to achieve a K^+ ion concentration of 0.5 mol.dm^-3?

4.2 You have a 500 ml container in which to prepare a solution of potassium permanganate (KMnO_4).

4.2.1 What mass of KMnO_4 is needed to make a solution with a concentration of 0,2 M?

4.2.2 Ignoring the container size, calculate the volume (in ml) of K_2 SO_4 that must be added to the KMnO_4 to achieve a K^+ ion concentration of 0,5 mol⋅dm^(-3).

### 2 Answers

- 6 months ago
@pisgahchemist Thank you for your answer. Checking with the examiner, we must make a 500 mL solution of KMnO4 and I did arrive to the same answer you did.

I agree with you with regards to the second part. Either the volume or concentration needs to be known.

Thanks again.

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- pisgahchemistLv 76 months ago
KMnO4 solution .....

You may have a 500 mL container, but do you want to make 500 mL of the 0.200M KMnO4 solution?

0.500L x (0.200 mol KMnO4 / 1 L) x (158.0g KMnO4 / 1 mol KMnO4) = 15.8g KMnO4 .... if we assume that you really meant to express the numbers to three significant digits, instead of one.

The second part is nonsense, since you must know the concentration of the K2SO4 solution which is to be poured into the KMnO4 solution.

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