Question

A student determined their NaOH molarity to be 0.0943 M. This solution was used to standardize 25.00 mL of an HCl solution, requiring 25.73 mL to reach the equivalence point. The HCl solution was then used to determine the weight % of soda ash in 0.5102 g of an unknown sample, requiring 38.77 mL. If the determined NaOH concentration was 5.000% too low (relative error of -5%), how does it change the reported value of weight % soda ash?

Answer 1

The reaction between NaOH and HCl is

NaOH+HCl--------àNaCl+ H2O

1 mole of NaOH hence requires 1 mole of HCl to produce 1 mole of NaCl.

Moles of NaOH (assuming a concentration of 0.0943M) is =molarity* Volume in L= 0.0943*25.73/1000

Let c= concentration of HCl, moles of HCl consumed =C*25/1000

Hence 0.0943*25.73/1000= C*25/1000

C= 0.0943*25.73/25=0.097054

When NaOH concentration determined was 5% too low,

C1= new HCl concentration was = 0.0943*25.73*0.95/25=0.0922M

The reaction between HCl and Na2CO3 in soda ash is

Na2CO3+2HCl ----à2NaCl+ H2CO3

1 mole of Na2CO3 requires 2 mole of HCl

Hence when C1 is the concentration of HCl

Moles of HCl in 38.77ml of HCl = 38.77*0.0922/1000 = 0.003575 moles

Moles of Na2CO3=0.5 times HCl =0.5*0.003575= 0.001787

Mass of Na2CO3= moles* molar mass= 0.001787*106=0.189 gm

Weight % of Na2CO3= 100*0.189/0.5102= 37.04%

When C is concentration of HCl

Moles of Na2CO3 =0.5*38.77*0.0977/1000=0.01894

Mass of Na2CO3= 0.01894*106= 0.200755 gm

Mass % Na2CO3= 100*0.20755/0.5102= 40.68%

% error =10*{ (40.68-37.04)/37.05 =9.8%