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Topic: pH  (Read 2646 times)

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Offline Arbillea

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pH
« on: September 19, 2010, 01:39:06 PM »
What is the pH of the resulting solution when 100.0 mL of 0,04M KMnO4 and 0,2M H2SO4 is mixed with 100.0 mL of 0.4M KI?



2KMnO4 + 8H2SO4 + 10 KI ----> 2MnSO4 + 8H2O + 5I2 + 6K2SO4

After reaction [H2SO4]= (0,02 - 0,016)/0,2 = 0,02M
[H3O+]= 2[H2SO4] = 0,04M

pH= -log(0,04)
pH= 1,4

Is this correct?

Offline sjb

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Re: pH
« Reply #1 on: September 19, 2010, 04:15:47 PM »
What is the pH of the resulting solution when 100.0 mL of 0,04M KMnO4 and 0,2M H2SO4 is mixed with 100.0 mL of 0.4M KI?



2KMnO4 + 8H2SO4 + 10 KI ----> 2MnSO4 + 8H2O + 5I2 + 6K2SO4

After reaction [H2SO4]= (0,02 - 0,016)/0,2 = 0,02M
[H3O+]= 2[H2SO4] = 0,04M

pH= -log(0,04)
pH= 1,4

Is this correct?

Right sort of idea, I think. But what volume of sulfuric acid did you start with? What is the limiting reagent? What is the new volume?

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