CHEM 11111
Calculations in Chemistry
Dr. S. Sri Skandaraja
Department of Chemistry
University of Kelaniya
For pH 5.21 ±0.03, find [H+] and its uncertainty.
pH = - log[H+]
log[H+] = - pH
10log[H+] = 10-pH
But, 10log[H+] = [H+]
You prepared a 0.250 M NH3 solution by diluting 8.45 (±0.04)
mL of 28.0 (±0.5) wt% NH3 [density 0.899 (±0.003) g/mL] up
to 500.0 (±0.2) mL. Find the uncertainty in 0.250 M. The
molecular mass of NH3, 17.030 5 g/mol, has negligible
uncertainty relative to other uncertainties in this problem.
Grams of NH3 per mL in concentrated reagent =
0.899 (g of solution / mL) x 0.280 (g NH3 / g solution)
= 0.2517 g of NH3 / mL
Moles of NH3 in 8.43 mL =
0.2517 (g of NH3 / mL) x 8.45 mL
17.0803 (g of NH3 / mol)
= 0.1249 mol
Molarity = 0.1249 mol / 0.5000 L
= 0.2498 mol / L
Grams of NH3 per mL in concentrated reagent
= 0.899 (±0.003) (g of solution / mL) x 0.280 (±0.005) (g NH3 / g solution)
= 0.899 (±0.334%) (g of solution / mL) x 0.280 (±1.79%) (g NH3 / g solution)
= 0.2517 (± 1.82%) g of NH3 / mL
Moles of NH3 in 8.43 mL =
0.2517(±1.82%) (g of NH3 / mL) x 8.45 (±0.473%) mL
17.0803 (±0%) (g of NH3 / mol)
= 0.1249 (±1.88%) mol
Find the uncertainty in [NH3] if the starting reagent
is 28.0 (±0.7) wt% NH3.
Systematic error
Systematic error occurs in some common
situations and is treated differently from
random error in arithmetic operations.
The atomic mass of oxygen as 15.999 4 ±
0.000 3 g/mol.
The uncertainty is not mainly from random
error in measuring the atomic mass.
The uncertainty is predominantly from isotopic
variation in samples of oxygen from different
sources.
The atomic mass of oxygen in a particular lot
of reagent has a systematic uncertainty.
It could be relatively constant at 15.999 7 or
15.999 1, or any value in between, with only a
small random variation around the mean value.
For systematic uncertainty, we add the
uncertainties of each term in a sum or
difference.
Differentiation (අවකලනය)
is an operation that allows us to find a
function that outputs the rate of change of
one variable with respect to another
variable.
y = mx + C
y2 - y1
m=
y
x2 - x1
(y2,x2)
y
m=
(y1,x1)
x
x
y = x2
y
(4,16)
(3.5,12.25)
(3,9)
x
y2 - y1
y
m=
x
=
x2 - x 1
=
16 - 9 - 9
12.25
43.5
- 3- 3
=7
6.5