Experiment 3:  Analysis of a Mixture of Carbonate and Bicarbonate

Introduction:

The purpose of this lab is to become familiar with the technique of titration.  Several titrations were performed to first standardize a base and then use the standardized base to standardize an acid.  The acid was used to titrate a solution to be able to determine the amount of carbonate and bicarbonate in an unknown.

Data:

 Unknown A amount of KHP used (g) Standardization of NaOH 1 .5112 g 1 25.9 ml 2 .5108 g 2 25.3 ml 3 .5102 g 3 25.6 ml Standardization of HCl: NaOH needed: HCl added to unknown 1 9.3 ml 1 75.3 ml 2 8.9 ml 2 75.2 ml 3 8.9 ml 3 75.5 ml HCl added to unknown, NaOH, and BaCl2 1 63.2 ml 2 63 ml 3 62.9 ml

Calculations:

Average for standardization of NaOH:  (25.9 + 25.3 + 25.6)/3= 25.6 ml +/-.3

Standard deviation:  [((25.9 – 25.6)2 + (25.3 – 25.6)2 + (25.6 – 25.6)2)/2]1/2= .3

Average for standardization of HCl:  (9.3 + 8.9 + 8.9)/3= 9.03 ml +/-.23

Average for HCl added to unknown:  (75.3 + 75.2 + 75.5)/3= 75.3 ml +/-.158

Average for HCl added to unknown, NaOH, and BaCl2:  (63.2 + 63.0 + 62.9)/3= 63.03 ml +/-.243

Solutions:

Amount of HCl needed to make 0.1 M HCl (1 L):

M1V1=M2V2

12.1 * V = .1 * 1         V= .00826 L = 8.26 ml HCl needed

Amount of NaOH needed to make 0.1 M NaOH (500 ml):

.1 M = x mol/.5 L

.5 L * .1 M=.05 mol * 39.998 g / 1 mol = 2 g NaOH needed

Amount of KHP needed:

25 ml * (.1 mol/1000 ml) * (204.234 g KHP/ 1 mol) = .5106 g KHP needed

Molarity of NaOH:

Trial 1:  .5112 g KHP * (1 mol KHP/204.233 g ) * (1 mol NaOH/ 1 mol KHP) * (1/.0259 L)= .09664 M

Trial 2:  .5108 g KHP * (1 mol KHP/204.233 g ) * (1 mol NaOH/ 1 mol KHP) * (1/.0253 L)= .09886 M

Trial 3:  .5102 g KHP * (1 mol KHP/204.233 g ) * (1 mol NaOH/ 1 mol KHP) * (1/.0256 L)= .09758 M

Average= .09769 M

Standard Deviation= .00111

Molarity of HCl:

Trial 1:  (.09769 M NaOH/1 L) * .0259 L NaOH * (1 mol HCl/1 mol NaOH) * (1/.0093 L HCl)= .272 M

Trial 2:  (.09769 M NaOH/1 L) * .0253 L NaOH * (1 mol HCl/1 mol NaOH) * (1/.0089 L HCl)= .278 M

Trial 3:  (.09769 M NaOH/1 L) * .0256 L NaOH * (1 mol HCl/1 mol NaOH) * (1/.0089 L HCl)= .281 M

Average= .277 M

Standard Deviation= .0046

Relative percent Carbonate and Bicarbonate:

Total Alkalinity= [HCO3-] + 2[CO32-]

HCO3- + 2H+ = H2CO3

CO32- + 2H+ = H2CO3

(.277 mol HCl/1 L) * .0753 L HCl * (1 mol HCO3-/1 mol HCl) * (1/.005 L unknown)= 4.17 M HCO3-

(.277 mol HCl/1 L) * .0753 L HCl * (1 mol CO32-/2 mol HCl) * (1/.005 L unknown)= 2.02 M HCO3-

Total Alkilinity= 4.17 + 2 * 2.02= 8.21 M

Mol HCO3-:

.05 L NaOH * .09769 M NaOH * (1 mol OH-/1 mol NaOH)= .004885 mol OH-

.06303 L HCl * .272 M HCl * (1 mol H+/1 mol HCl)= .01714 mol H+

.004885 mol OH- - .01714 mol H+ = -.0122 mol HCO3-

This number is negative and therefore the percent carbonate and bicarbonate cannot be calculated.

Conclusions:

The main goal of calculating the percent carbonate and bicarbonate was not achieved because a negative number of moles of HCO3- were calculated.  This was the result of a large error in the experiment.  Part of the error occurred because the unknown was a liquid and not a solid.