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Topic: Determination of an unknown concentration of Hydrochloric acid  (Read 5999 times)

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Offline skeme

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Determination of an unknown concentration of Hydrochloric acid
« on: October 01, 2006, 07:12:22 PM »
I was hoping that someone could check the following calculations and also help me out solving 2 other problems

molarity of potassium hydrogen phthalate

5.1658/(204*0.25) = 0.1012

moles of potassium hydrogen phthalate

0.1012*0.025 = 0.0253

molarity of NaOH

0.0253/0.0253 = 1 mol

Calculation of the unknown HCL concentration

the number of moles of NaOH involved in the reaction

0.0253/0.0263 = 0.96

Now i need to calculate the number of moles of HCL involved in the reaction and the molarity of the HCL but I'm totally confused



« Last Edit: October 01, 2006, 08:00:40 PM by skeme »

Offline enahs

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Re: Determination of an unknown concentration of Hydrochloric acid
« Reply #1 on: October 01, 2006, 08:21:03 PM »
You need to be a little clearer. You just spout out random calculations. We can not tell you if they are correct or not unless we know what we are doing.  I mean sure, this is correct mathematically [5.1658/(204*0.25) = 0.1012 ], but it is just random numbers and not possible to tell you if it is correct chemically.


What do you know and what are you trying to get, then tell us what you have done so far and we can check that work.

Offline skeme

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Re: Determination of an unknown concentration of Hydrochloric acid
« Reply #2 on: October 01, 2006, 08:57:29 PM »
Sorry if this doesn't make anymore sense than my first post it's just gone 2am here.

The first part of the experiement was to make a standard out of potassium hydrogen phthalate. Then after finding out the concentration of NaOH i titrated an unknown concentration of Hydrochloric acid.

I have worked out the molarity of the potassium hydrogen phthalate

Mass of potassium hydrogen phthalate
molar mass*volume (made up to 250ml)

5.1658/(204*0.25) = 0.1012

after that i used the answer to work out the moles of potassium hydrogen phthalate by multiplying the molarity of potassium hydrogen phthalate by the aliquot size (25ml) to give me

 0.1012*0.025 = 0.0253  not sure if this is correct

then i calculated the molarity of the NaOH i used by dividing the molarity by my titre

0.0253/0.0253 = 1 mol

Next i need to work out the number of moles of NaOH involved in the reaction

i did ths by dividing the moles of NaOH by my titre and got

0.0253/0.0263 = 0.96


Now i need to calculate the number of moles of HCL involved in the reaction
and the molarity of the HCL

Offline enahs

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Re: Determination of an unknown concentration of Hydrochloric acid
« Reply #3 on: October 01, 2006, 10:16:12 PM »
Quote
0.1012*0.025 = 0.0253  not sure if this is correct
No, it is not. It is 0.00253

This # is how many moles of KHP where in 25 mL.

You next steps are confusing.

KHP and NaOH react together 1:1
So you know if it took 25 mL of KHP (and thus 0.00253 moles) to neutralize your NaOH you have that many moles of NaOH. Just divide that number by the volume.

Say the 25mL of KHP neutralized 50 mL of NaOH, that means the Molarity of your NaOH is 0.00253 moles/ 0.050L, so if this was the correct amount of NaOH the 25mL of KHP neutralized then you would have 0.05M NaOH.

Your NaOH also reacts 1:1 with the HCl.

So your x liters of NaOH, xL * Molarity (0.00253M) = moles of NaOH required to titrate HCl = Moles of HCl.

Once you know moles of HCl, then you just have to divide it by the volume of HCl titrated to get molarity.

Offline skeme

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Re: Determination of an unknown concentration of Hydrochloric acid
« Reply #4 on: October 03, 2006, 01:30:00 PM »
Thanks for the help enahs  ;D

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