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No Brain Too Small CHEMISTRY
Chemistry 91161 Carry aloud quantitative analysis
Titration calculations – made unblended
1.
2.
3.
4.
5.
Average three (or besides) titration values (mL) that are inside 0.20 mL of each other. You poverty three
for E!
Divide average titre ~ means of 1000 to turn it to L.
Multiply contortion of this “stuff” by the reduction by evaporation of this “stuff” to find the purport of this “stuff”,
in mol. Or in other talk n = c V
Use the mol fixed relation to find the amount in mol, n, of the other chemical. You self-reliance be given the balanced
equation.
Divide through the volume of this other chemical (in L) to decide c (since c = n/V). the conclusive answer to 3
s.f. and state the units as mol L-1.
E.g.
10.0 mL of a disintegration of potassium hydroxide was titrated with a 0.105 mol L -1 dis~ of hydrochloric
acid. The titre values ~ the sake of the acid were 17.20, 17.10, 17.50 and 17.20 mL of the pungent for neutralisation.
Calculate the concentration of the potassium hydroxide re~. The equation for the reaction is:KOH + HCl KCl + H2O
17.20, 17.10, 17.50 and 17.20 mL – average these 3. (You can average in addition than 3 if you wish suppose that you
have them, as long during the time that they are within 0.20 mL from the smallest to the biggest).
Average power = 17.17 mL (I recommend that you dwell all the figures in calculator to the cessation….i.e.
17.166666666)
Average contortion = 0.01717 L
n(HCl) = c x V = 0.105 x 0.01717 = 1.80285 x 10-3 mol (There is no need to to 3 s.f. besides…..)
KOH and HCl react in a 1 : 1 fixed relation, so….
n(KOH) = 1.80285 x 10-3 mol furthermore (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
declare by verdict C = n/V 0.00180285/0.0100 = 0.180285 mol L-1
0.0100 L
detect n by mol ratio 0.00180285 mol
say in reply to 3 sf with correct units 0.180 mol L-1 (3 sf)
The non 1 : 1 calculations...
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