 No Brain Too Small  CHEMISTRY 

Chemistry 91161 Carry out quantitative analysis
Titration calculations – made simple
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Average three (or more) titration values (mL) that are within 0.20 mL of each other. You need three for E!
Divide average titre by 1000 to turn it to L.
Multiply volume of this “stuff” by the concentration of this “stuff” to find the amount of this “stuff”, in mol. Or in other words n = c V
Use the mol ratio to find the amount in mol, n, of the other chemical. You will be given the balanced equation. Divide by the volume of this other chemical (in L) to find c (since c = n/V). Write the final answer to 3
s.f. and put the units as mol L-1.

E.g.
10.0 mL of a solution of potassium hydroxide was titrated with a 0.105 mol L -1 solution of hydrochloric acid. The titre values for the acid were 17.20, 17.10, 17.50 and 17.20 mL of the acid for neutralisation.
Calculate the concentration of the potassium hydroxide solution. The equation for the reaction is:KOH + HCl  KCl + H2O
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17.20, 17.10, 17.50 and 17.20 mL – average these 3. (You can average more than 3 if you wish if you have them, as long as they are within 0.20 mL from the smallest to the biggest).
Average volume = 17.17 mL (I recommend that you keep all the figures in calculator to the end….i.e.
17.166666666)
Average volume = 0.01717 L n(HCl) = c x V = 0.105 x 0.01717 = 1.80285 x 10-3 mol (There is no need to write to 3 s.f. yet…..)
KOH and HCl react in a 1 : 1 ratio, so…. n(KOH) = 1.80285 x 10-3 mol also (still keep the number in your calculator!!)
Remember 10.0 mL of KOH is 0.0100 L c(KOH) = n/V = 1.80285 x 10-3 / 0.0100 = 0.180285 = 0.180 mol L-1 (3 sf)

c
V
n

HCl
0.105 mol L-1
0.01717 L
 n = CV 0.00180285 mol

KOH
 find C = n/V 0.00180285/0.0100 = 0.180285 mol L-1
0.0100 L
 find n by mol ratio 0.00180285 mol

 Write answer to 3 sf with correct units 0.180 mol L-1 (3 sf)
The non 1 : 1 calculations will either be 2 : 1 or 1: 2 ratios. So at step 2 you will need to either x2 or 2.
Which one depends on what you need to find e.g.
Na2CO3 + 2HCl  etc

If you know n(Na2CO3) and need to find n(HCl), then x2
If you know n(HCl) and need to find n(Na2CO3), then 2

 No Brain Too Small  CHEMISTRY 
LET’S LOOK AT IT ANOTHER WAY….
It’s as easy as n, n, c (okay it’s not as easy as A, B, C but it’s close enough).
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Find n of the stuff you know 2 things about (its c and V)
Find n of the other stuff, taking into account the mole ratio
Find c of this other stuff, since you now have V and n
Write it to 3 s.f. and put the units mol L-1 on it and you are done!! (Watch out you don’t write the units as mol-1 or moL-1!!! Trust me – we have seen it happen and it’s WRONG!!!)

Sample calculation: Standardising a solution of sodium hydroxide using a standard solution of hydrochloric acid. Titrate 20.0 mL aliquots of sodium hydroxide solution using 0.0950 mol L -1 hydrochloric acid.
NaOH + HCl  NaCl + H2O
Rough
Final volume
(mL)
Initial volume
(mL)
Volume used
(mL)

1

2

3

4

21.16

41.21

20.75

40.95

20.50

0.10

21.16

0.60

20.75

0.30

21.06

20.05

20.15

20.20

20.20

Select at least three concordant results:
You could select 20.05 mL and 20.15 mL and 20.20 mL and 20.20 mL, or you might decide to just use the last three (20.15 mL, 20.20 mL and 20.20 mL).
Alternatively this student could have chosen to stop after titration 3 as 20.05 mL, 20.15 mL and 20.20 mL are all within 0.2 mL of each other.
The student could even use 20.05 mL and 20.20 mL and 20.20 mL but you have to wonder why they would choose to discard the 20.15 mL.
You must show which values you have selected.
V(HCl) = (20.05 + 20.15 + 20.20 + 20.20) / 4
V(HCl) = 20.15 mL = 0.02015 L
1. n(HCl) = cV = 0.0950 x 0.02015 = 0.00191425 mol
2. n(NaOH) = 0.00191425 mol
3. c(NaOH) = n/V = 0.00191425 / 0.0200 = 0.0957125 so c(NaOH)