Question

A volume of 25.06 ± 0.06 mL of HNO3 solution was required for complete reaction with 0.9781 ± 0.0007 g of Na2CO3, (FM 105.988 ± 0.001). Find the molarity of the HNO3 and its absolute uncertainty.

Answer 1

Reaction:Na2CO3+2HNO3=2NaNO3+CO2+H2O

Mass of Na2CO3 reacted=0.9781 ± 0.000g

Moles of Na2CO3 reacted=0.9781 ± 0.000g/molar mass of Na2CO3

molar mass of Na2CO3=23*2+12+16*3=105.988 +/-0.001 g/mol

Moles of Na2CO3 reacted=0.9781 ± 0.0007g/(105.988 +/-0.001 g/mol)

Using rule for propagation of errors of multiplication/division,

R=X/Y

∆R=lRl*[(∆X/X)^2+(∆Y/Y)^2]^1/2

∆R=0.00923*[(0.0007/0.9781)^2+(0.001/105.988)^2]^1/2=0.00923*[0.000000511+0.000000000088]^1/2=0.0000066

Moles of Na2CO3 reacted=(0.9781/105.988) =0.00923 +/- 0.0000066 moles

As 1 mole Na2CO3 reacts with 2 moles of HNO3

So 0.00923 +/- 0.0000066 moles of Na2CO3 reacts with 2 *(0.00923 +/- 0.0000066) moles of HNO3=0.0184+/-0.0000132 moles

Molarity of HNO3=moles/volume=(0.0184+/-0.0000132 moles)/( 25.06 ± 0.06 mL)

Using rule for propagation of errors of multiplication/division,

R=X/Y

∆R=lRl*[(∆X/X)^2+(∆Y/Y)^2]^1/2

Molarity of HNO3=moles/volume=(0.0184+/-0.0000132 moles)/( 25.06 ± 0.06 mL)=0.000734+/- ∆M

∆M=0.000734*[(0.0000132/0.0184)^2+(0.06/25.06)^2]^1/2=0.000734[0.00000052+0.0000057]^1/2

=0.00000183

Molarity of HNO3=0.000734+/- 0.00000183 mol/ml * 1000 ml/L=(0.000734*1000)+/- (0.00000183*1000)=0.734*0.00183 mol/L=0.734*0.00183M(answer)