Question 1: A solution of F– is prepared by dissolving 0.0712 ± 0.0005 g of NaF

Question

Question 1: A solution of F– is prepared by dissolving 0.0712 ± 0.0005 g of NaF (molecular weight = 41.989 ± 0.001 g/mol) in 164.00 ± 0.07 mL of water. Calculate the concentration of F– in solution and its absolute uncertainty.

Question 20. A solution of HNO3 is standardized by reaction with pure sodium carbonate.

A volume of 27.64 ± 0.05 mL of HNO3 solution was required for complete reaction with 0.9875 ± 0.0009 g of Na2CO3, (FM 105.988 ± 0.001). Find the molarity of the HNO3 and its absolute uncertainty.

Explanation / Answer

1)

The concentration of NaF = Mass of NaF X 1000 / Molecular weight of NaF X volume of water

concentration of NaF = 0.0712 X 1000 / 41.989 X 164 = 0.013 M

Now uncertainities in multiplication and division will propagate as

relative uncertainity in concentration

= [(relative uncertainity in mass)2 + (relative uncertainity in molecular weight)2 + (relative uncertaintiy in volume)2] 1/2

Relative uncertainity in concentration = [(0.0005 / 0.0712)2 + (0.001/41.989)2 + (0.07/164)2]1/2

Relative uncertainity = [ (0.000049 + 5.66X10^-10 + 18.14X10^-8 ) 1/2 = 0.007

Absolute uncertainity in concentration = Relative uncertainity X actual value = 0.007 X 0.013 = 0.000091

2) the reaction of acid with the given base will be

2HNO3 + Na2CO3 --> 2NaNO3 + H2O + CO2

As per equation the moles of HNO3 will be twice that used of Na2CO3

(Molarity X volume ) HNO3 = 2X (moles) Na2CO3

Moles of Na2CO3 = Mass / Molecular weight = 0.9875 / 105.988 = 0.00932

relative uncertainity in moles of Na2CO3 = [(0.0009/0.9875)2 + (0.001/105.988)2]1/2

Relative uncertainity in moles of Na2CO3 = (83.046 X 10^-8 + 89.019 X 10^-12]1/2

Relative uncertainity in moles of Na2CO3 = 0.000911

(Molarity X volume ) HNO3 = 2X (moles) Na2CO3 = 2 X 0.00932

molarity = 0.674 Molar

Relative uncertainity in molarity = [relative uncertaintiy in moles )2 + (relative uncertainity in volume)2]1/2

Relative uncertainity in molarity = [(0.000911)2 + (0.05/27.64)2 )1/2

Relative uncertainity in molarity =    [8.2X10^-7 +3.24 X 10^-6]1/2 = 0.002

So absolute uncertainity in molarity = Relative uncertainity in molarity X molarity = 0.002 X 0.674 = 0.00135

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