We can measure the concentration of HCl solution by its reaction with pure sodiu

Question

We can measure the concentration of HCl solution by its reaction with pure sodium carbonate.

Complete reaction with 0.9639 ± 0.0005 g of Na2CO3 required 28.20 ± 0.05 mL of HCl.

(a) Find the formula mass (and its uncertainty) for Na2CO3.

(b) Find the molarity of the HCl and its absolute uncertainty. (Enter an unrounded value. Use at least one more digit than given.)

(c) The purity of primary standard Na2CO3 is stated to be 99.95 to 100.05 wt%, which means that it can react with(100.00 ± 0.05)% of the theoretical amount of H+. Recalculate your answer to (b) with this additional uncertainty. (Enter an unrounded value. Use at least one more digit than given.)

Explanation / Answer

a) formula mass of Na2CO3=23*2+12+16*3=106 g/mol

b) Mass= 0.9639 ± 0.0005 g of Na2CO3 required 28.20 ± 0.05 mL of HCl.

Molar mass of Na2CO3=23*2+12+16*3=106 g/mol

Moles of Na2CO3=mass/molar mass== (0.9639 ± 0.0005 g )/106 g/mol=0.00909 +/- 0.00000471 moles

Rule for propagation of error ,multiply by constant,

X*constant +/- X

Where X=constant*lXl

From the reaction ,2 moles of HCl reacts with one mole Na2CO3

2 HCl + Na2CO3 2 Na+ + H2O + CO2

So 0.00909 +/- 0.00000471 moles of Na2CO3 reacts with 2(0.00909 +/- 0.00000471 moles

)=0.01818+/- 0.00000942 moles of Hcl

Molarity =Concentration of HCl=moles/volume=(0.01818+/- 0.00000942 moles )/( 28.20 ± 0.05 mL)

                                                                 =0.000645 +/- 0.002288 mol/ml *1000 ml/L

                                                                  =0.645+/- 0.0012 mol/L

[[(use Rule for propagation of error in multiplicqation R=X*Y

R=lRl*[(X/X)^2 +(Y/Y)^2]^1/2=0.000645 *[(0.00000942/0.01818)^2+(0.05/28.20)^2]^1/2=0.002288

R = 0.645 * [(0.00000027 +0.0000031]^1/.2=0.0012]]

c) Theoretical amount of Hcl (calculated in part b)= 0.01818+/- 0.00000942 moles of Hcl

theoretical mass of HCl with which the given amount of Na2CO3 can react=0.01818+/- 0.00000942 moles of Hcl *molar mass of Hcl (36.5 g/mol)=0.6636+/- 0.000344 g

But Na2Co3 can react with (100.00 ± 0.05)% of the theoretical amount of HCl

Or, Na2Co3 can react with (100.00 ± 0.05)/100 *(0.6636+/- 0.000344 g)=(1+/-0.0005)*( 0.6636+/- 0.000344 g)=(0.6636 +/- R)

R=lRl*[(X/X)^2 +(Y/Y)^2]^1/2 where X=1+/-0.0005 ,Y=( 0.6636+/- 0.000344 g)

R=0.6636*[(0.0005/1)^2+(0.000344/0.6636)^2]^1/2=0.6636*[0.00000025+0.00000027)^1/2=0.000720

So R=0.000720

Therefore, Na2Co3 can react with =(0.6636 +/- R)= 0.6636 +/- 0.000720 g of Hcl

Or Na2Co3 can react with Moles of Hcl =0.6636 +/- 0.000720 g of Hcl * molar mass of Hcl(36.5 g/mol)=0.01818+/- 0.0000197 moles of Hcl

Molarity of Hcl=moles of Hcl/volume of Hcl=(0.01818+/- 0.0000197 moles)/ ( 28.20 ± 0.05 mL)                                                                                              =(0.01818+/- 0.0000197 moles)/ ( 28.20 ± 0.05 mL)

=0.000645+/- Z

Z=0.000645*[(0.0000197/0.01818)^2 +(0.05/28.20)^2]^1/2=0.000645*[0.0000000117+0.00000313]^1/2=0.00000114

Molarity of Hcl=(0.000645+/- 0.00000114 g/ml) *1000 ml/L=0.645+/-0.00114 mol/L(answer)

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.