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### Topic: Stoichiometry Concentration Problem  (Read 5008 times)

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#### positiveion

• Regular Member
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• Mole Snacks: +2/-7 ##### Stoichiometry Concentration Problem
« on: September 22, 2009, 11:12:23 PM »
In a titration, 25cm^3 of .1M KOH is pipetted into a conical flask. Sulphuric acid is run in from the burette and it was found that 19.8cm^3 were required to react exactly. CAclulate the concentration of the H2SO4.

what i did was:

KOH : H2SO4
0.0025g:Xg
56.11g:98.08

^ 0.0025 is the no. grams of KOH dissolved in the solution. 56.11 and 98.08 are the mass of 1 mole of each substance

So using those ratios:

x/0.0025 = 98.08/56.11
x=0.00436...grams

That's the no. grams in H2SO4

To find the no. moles I multiply it by 98.08, which equals .4286083..

Concentration = no. moles / volume

So

.4286../19.8 = .4286083764 M

But the correct answer is supposed to be 0.063M and I don't know where I went wrong

#### UG

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• Mole Snacks: +134/-15
• Gender:  ##### Re: Stoichiometry Concentration Problem
« Reply #1 on: September 22, 2009, 11:36:30 PM »
What you need first is a balanced equation.
You'll find that it is a 2:1 ratio of base : acid since sulfuric acid donates 2 moles of H+ for every mole of acid. You find the number of moles of KOH, n= conc. x volume. The number of moles of sulfuric acid will be half of this, you then find the concentration of the acid using conc. = n/volume

#### positiveion

• Regular Member
•   • Posts: 79
• Mole Snacks: +2/-7 ##### Re: Stoichiometry Concentration Problem
« Reply #2 on: September 23, 2009, 12:25:50 AM »
Oh -headdesk- Okay that was a stupid error, but um I'm trying to do it using ratios so could you tell me where I went wrong in the following:

2KOH + H2SO4
.0025 : x
112.22:98.08

(^ .0025 is the KOH in the 25 cm cubed solution, 112.22 is 2KOH and 98.08 is H2SO4)

x/.0025 = 98.08/112.22
x=.0021849938
^ thats the number of grams in H2SO4

.0021849938/98.08 = 2.2277 x 10^-5
^ I put the no. grams over the no. grams in one mole to find the answer of how many moles there are

Now that I have that number of moles I put it in the the eqution

Concentration = (2.2277x10^-5)/19.8
Concentration = 1.125 x 10^-6

>_<

How did I mess up?

#### Borek ##### Re: Stoichiometry Concentration Problem
« Reply #3 on: September 23, 2009, 03:04:37 AM »
You don't need masses for that, do all your calculations using moles.

What is 0.0025?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### positiveion

• Regular Member
•   • Posts: 79
• Mole Snacks: +2/-7 ##### Re: Stoichiometry Concentration Problem
« Reply #4 on: September 23, 2009, 05:23:46 AM »
That's the mass of KOH.

1000*.025=25 ml
.5 * .025 = .0025g

#### Borek ##### Re: Stoichiometry Concentration Problem
« Reply #5 on: September 23, 2009, 05:38:58 AM »
No idea what you are doing, the only thing I am sure about is that it is wrong. 0.0025 is not a mass of KOH, unless you are using some exotic mass units.

What is 0.5?

But it doesn't matter - as I wrote earlier, forget about converting to masses. It doesn't make sense to convert moles to mass, to calculate mass of sulfuric acid from mass of KOH and then to convert mass back to moles, if you can calculate moles of sulfuirc acid directly from moles of KOH.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info