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LEASE REMEMBER TO USE YOUR’S SCHOOL’S AVERAGE TITRE VALUE(AVERAGE VOLUME OF ACID). WE USED 15.26cm3, TRY TO KNOW YOUR SCHOOL’S AVERAGE TITRE VALUE. ASK YOUR CHEMISTRY TEACHER. THEN, ANYWHERE YOU SEE 15.26cm3 IN MY CALCULATION, PUT YOUR SCHOOL’S OWN AND RE-CALCULATE. THIS IS VERY IMPORTANT!!!
(1a)
In a tabular form
Burette Reading|1st reading|2nd reading|3rd reading
Final |15.25|0.00|45.79|
Initial|0.00||15.25|30.53|
Volume of acid used |15.25|15.28|15.26
Average volume of acid used = 15.25 + 15.26/2
= 15.255cm³
=15.26cm³
(1bi)
Given: Mass of conc of A = 5g/500cm³ = 5g/0.5dm³
Ca = 10g/dm³
A is HNO3
Therefore; Molar mass = 1+14+(16×3)
= 15+48
=63g/mol
Molarity of A = gram conc/molar mass
Ca = 10/63 = 0.1587mol/dm³
(1bii)
Using CaVa/CbVb = nA/nB
With reacting equation:
HNO3 +NaOH–>NaNO3 + H2O
nA = 1, nB = 1
0.1587×15.26/Cb×25.00 = 1/1
25Cb = 0.1587×15.26
CB = 0.1587×15.26/25
CB = 0.09687mol/dm³
(1biii)
B is NaOH
Molar mass = 23+16+1
=40g/mol
Conc of B in g/dm³ = molarity×molar mass
=0.09687×40
=3.8748g/dm³
(1biv)
No of moles present in 250cm³ of NaOH is
= molar conc. × volume
= 0.09687 × 250/1000
= 0.0242 moles
Mole ratio of NaOH and NaNO3 is 1:1
No of moles of NaNO3 which reacted is 0.0242
Mass of NaNO3 formed = molar mass × no of moles
= 85 × 0.0242
=2.057grams
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