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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: peterschmidt3943 on March 13, 2018, 05:54:32 PM

Title: Molarity
Post by: peterschmidt3943 on March 13, 2018, 05:54:32 PM
Hello,

why does 0,005 M of H2SO4 have 0,01 M of protones?
Title: Re: Molarity
Post by: sjb on March 13, 2018, 06:08:01 PM
Why do you think it doesn't? What is a proton?
Title: Re: Molarity
Post by: peterschmidt3943 on March 13, 2018, 06:11:31 PM
Well, 0,005 M already describes how many moles it got per litre...

I dont understand why one has to calculate 0,005*2 in order to get to the moles/l of the protones in the molceule...
Title: Re: Molarity
Post by: peterschmidt3943 on March 13, 2018, 06:20:27 PM
or to put it in different words: Why do I have to calculate the number of protones in a neutralisation reaction in oder to see to what extent it reacted with the base?? Why not calculate with 0,005 M?
Title: Re: Molarity
Post by: Borek on March 13, 2018, 06:27:34 PM
Well, 0,005 M already describes how many moles it got per litre...

Moles of what?
Title: Re: Molarity
Post by: peterschmidt3943 on March 13, 2018, 06:34:28 PM
H2SO4
Title: Re: Molarity
Post by: Arkcon on March 13, 2018, 08:29:48 PM
But you have  0,005 M of H2SO4, how come you seem to have twice as many molar of protons?  Did you double them somehow?  Look at the formula.
Title: Re: Molarity
Post by: peterschmidt3943 on March 13, 2018, 08:56:15 PM
Hm? What do you mean exactly ?
Title: Re: Molarity
Post by: jestearns on March 14, 2018, 12:25:44 AM
Your question involves stoichiometry.  Each molecule of sulfuric acid is diprotic (has 2 hydrogen atoms per molecule).  Therefore,

0.005 mol H2SO4/L * 2 mol H+/mol H2SO4 = 0.01 mol H+/L
Title: Re: Molarity
Post by: peterschmidt3943 on March 14, 2018, 02:49:30 AM
is that the reason why a puffer based on 

10 ml of   0,1 M Phosphatpuffer and

100 ml of a 0,01 M Phosphatpuffer

consisting of the same amount of KH2PO4 und K2HPO4

can puffer the same amout of OH as well as H+ (because we would have 0,1*0,1 = 0,01 M protones in the first one; and since we got PO4 in the same amount, we have to calculate nothing here)

Why do they both have the same amount of electrolytes and difer in the concentration of the electrolytes?

Title: Re: Molarity
Post by: peterschmidt3943 on March 14, 2018, 03:25:51 AM
and the reason why I was so confused about my initial question was that my prof calculated M of H+ * V =  0,1M of NaOH * V
in order to see if we got equimolar compounds... why didnt he go with 0,005 of the whole compund H2PO4? Why did he not calculate by the same token the M of OH we got in NaOH??
Title: Re: Molarity
Post by: Borek on March 14, 2018, 03:50:35 AM
H2SO4

And how many moles of H+ per mole of H2SO4?
Title: Re: Molarity
Post by: peterschmidt3943 on March 14, 2018, 09:12:00 AM
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