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A researcher prepares serial dilutions of 3.2 M potassium hydroxide. He adds 1 mL KOH to...

A researcher prepares serial dilutions of 3.2 M potassium hydroxide. He adds 1 mL KOH to 9 mL of water in tube #1. He repeats this 4 more times. What is the concentration of KOH in the final dilution (tube #5)?

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Homework Answers

Answer #1

So, the final concentration of KOH in tube 5 is 32uM.

Second method -

Overall dilution is 10*10*10*10*10 = 105 times

1 M = 1000mM

1mM = 1000uM, So 1M = 1000000 uM

So, final concentration = 3.2M /100000 = 3200000uM/ 100000 = 32uM.

So, the concentration of KOH in 5th tube is 32uM.

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