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Author Topic: titration of HCl and H3PO4 mixture  (Read 10584 times)

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adrian_01

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titration of HCl and H3PO4 mixture
« on: April 11, 2004, 07:11:35 PM »

 Hello everyone...it is my first time using this site and I would like to ask for help with the folowing problem:

A 40.0 ml sample of a mixture of HCl and H3PO4 is titrated with 0.100 NaOH. The first equivalence point was reached after 88.0 ml of base, and the second equivalence point after 124.6 ml of base.
a) Find initial concetration of HCl and H3PO4 in solution.
b) What %of HCl is neutralized at first equivalence point.
c) What is the concetration of H3O+ at first equivalnece point.

          Thank you sooo much for your help and I hope I'll be able to answer in my turn other questions.
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AWK

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Re:titration of HCl and H3PO4 mixture
« Reply #1 on: April 12, 2004, 11:50:21 PM »

a) First equivalent point
HCl + NaOH = NaCl + H2O
H3PO4 + NaOH = NaH2PO4 + H2O
Second equivalent point
NaH2PO4 + NaOH = Na2HPO4 + H2O
c_molar
40xcHCl=(126.4-2x38.4)x0.100
40xcH3PO4=(38.4/3)x0.100

b) 100%
c) Calculate hydrolysis of 0.00384 mole of NaH2PO4 in volume (40+88) mL
« Last Edit: April 13, 2004, 07:10:16 PM by AWK »
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Dolphinsiu

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Re: titration of HCl and H3PO4 mixture
« Reply #2 on: December 19, 2006, 05:53:14 AM »

It seems your calculation is not correct! ;D

(a) no. of mol of Na2HPO4= [(124.6 - 88)/1000] x 0.1 = 3.66 x 10-3 mol

no. of mol of NaCl = 8.8 x 10-3- 3.66 x 10-3 = 5.14 x 10-3 mol

Then initial Conc. of H3PO4 = 3.66 x 10-3/0.04

Then initial Conc. of HCl =  5.14 x 10-3/0.04

(c) Calculate hydrolysis of 0.00366 mole of NaH2PO4 in volume (40+88) mL
« Last Edit: December 19, 2006, 06:50:17 AM by Dolphinsiu »
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AWK

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Re: titration of HCl and H3PO4 mixture
« Reply #3 on: December 19, 2006, 09:22:07 PM »

You are right. This small error comes from using 126,4 instead of 124,6.
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