Another successive dilution problem

  • Thread starter Benzoate
  • Start date
  • #1
421
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Homework Statement




The concentration of HCl in a bottle was determined as follows:

* 4.00 mL of the acid were diluted to 150.00 mL
* 26.53 mL of the diluted acid were titrated with 0.08329-M NaOH
* 43.91 mL of the base were required to reach the end point

What is the HCl concentration in the bottle?

Homework Equations



Possibly C(final)=C(initial)*(v(1)/v(2))*(v(3)/v(4))* ,....

The Attempt at a Solution



C(initial)= .08329 M
For end Volume I assume that be V(4) and V(3) would be the titrated volume.
C(final)=.08329 M *(4 mL/150 mL)*(26.53 mL/43.91mL)= .00134 M
 

Answers and Replies

  • #2
106
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2HCl + 2NaOH - > 2NaCl +2H20

Valence factor of NaOH = 1
Valence factor of HCl = 1

Normality of NaOH = 0.08329N
Equivalents used = 0.08329 * 43.91 = 3.657meq

SO equivalents of HCl in 26.53ml(dil) = 3.657meq

Eq in 154 ml water diluted sol = ???(Get the result in meq)

Volume of sol, = 154ML

U have got the meq U have got the Vol.U can find concentration in Normality or Molarity.
 
  • #3
106
0
Is the answer .77N right?
 

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