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nevada nevada
wrote...
Posts: 116
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10 years ago
i  have a test coming up on titration and i do not know or understand how to do any titration problems.


can you please explain the step needed to do titration problems

13. Consider the titration of 500.0 mL of 0.200 M NaOH with 0.800 M HCl. How many milliliters of 0.800 M HCl must be added to reach a pH of 13.000?




thank you.
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wrote...
Staff Member
Educator
10 years ago
pH=13

pOH =1

[OH] = 0.1 M

Let x ml of HCL be added

So according to question we have

0.1 = (0.2 *500 - 0.8*x) / (500 + x)

50 + 0.1x=100 -0.8x

0.9x =50

x=55.55 ml
Mastering in Nutritional Biology
Tralalalala Slight Smile
wrote...
10 years ago
as the reaction is
NaOH + HCL Rightwards Arrow NaCL + H20
let volume of HCl = x

moles of NaOH =0.200*0.500Ltr
= 0.1mol
moles of HCl =x*0.800 = 0.8x mol
stoichiometric balance:
NaOH + HCL Rightwards Arrow NaCL + H20
(0.1) (0.8x) (0.1) (0.1)
(0) 0.8x-0.1 (0.1) (0.1)
as now NaOH, H20, NaCL does not affect pH sa Na+ is conjugate of strong base NaOH and Cl- is conjugate to strong acid HCl.
and pH =13
[H+] =10^-13
so [H+] = (0.8*x) -0.1 = 10^-13
x =0.125ltr
=125mL
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nevada Author
wrote...
10 years ago
my teacher said that " titration of strong base vs strong acid, if partially titrated , has to be done using moles"  i don't really understand, how would i know when to use moles vs molarity?

it is like when i have a strong base and strong acid thats when i use moles

and if i should have a weak base and weak acid or weak acid with strong base i use molarity?
wrote...
Staff Member
Educator
10 years ago
molarity

Molarity (M) is the concentration of a solution expressed as the number of moles of solute per liter of solution.

So, if you know the M, you can figure out the moles, by multiplying that number by the volume (in litres).
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