Analysis of a Mixture of Carbonate and Bicarbonate
Purpose: The purpose of this lab is to use indirect titration to find the amount of Carbonate and
Bicarbonate in an unknown. Use titrations and standards.
Data:
Titration
NaOH
KHP
Titration
HCL
NaOH
KHP Grams
start mL
end mL
NaOH used
Moles NaOH
start mL
end mL
HCL used
Moles HCL
1
0.5111
0.2
27.5
27.3
0.091673743
1
5.6
32.8
27.2
0.088261864
2
0.5103
0.3
25.9
25.6
0.097608431
2
3.2
29.8
26.6
0.090252734
3
0.5105
0.5
25.8
25.3
0.098804552
3
2.9
29.4
26.5
0.09059331
A
0.510633333
Average
26.06666667
0.096028909
Average
26.76666667
0.089702636
Titration
HCL
Unknown
Titration
HCL
mixture
start mL
end mL
start
mL
end mL
HCL used
start mL
end mL
HCL used
1
1.3
50
2.3
10.5
56.9
1.5
31.2
29.7
2
10.5
50
5
21.2
55.7
1.8
31.5
29.7
3
21.2
50
2.6
19.7
45.9
1.8
31.7
29.9
Average
52.83333333
Average
29.76666667
mol of C/BiC
NaOH/Initial
NaOH/Excess
mols reacted
w/BiC
1
0.00510408
1
0.004801445
0.002664168
0.002137277
2
0.004996437
2
0.004801445
0.002664168
0.002137277
3
0.004117351
3
0.004801445
0.002682109
0.002119337
mol Carb
Rel % Carb
Rel % Bicarb
mol Bicarb
1
0.002137277
1
0.002966803
1
7.1212
1
5.2162
2
0.002137277
2
0.00285916
2
6.8628
2
5.2162
3
0.002119337
3
0.001998014
2
4.7958
3
5.1724
Aev.
6.2599
Aev.
5.2016
Stand Dev:
1.27453
Stand
Dev:
0.02528
Calculations:
. 025 𝐿 𝑡𝑖𝑡𝑟𝑎𝑛𝑡 ×
. 1 𝑚𝑜𝑙 𝑡𝑖𝑡𝑟𝑎𝑛𝑡
1 𝑚𝑜𝑙 𝐾𝐻𝑃
204.22 𝑔 𝐾𝐻𝑃
×
×
= 0.51 𝑔 𝐾𝐻𝑃
1 𝐿 𝑡𝑖𝑡𝑟𝑎𝑛𝑡
1 𝑚𝑜𝑙 𝑡𝑖𝑡𝑟𝑎𝑛𝑡
1 𝑚𝑜𝑙 𝐾𝐻𝑃
0.1 𝑀 𝑁𝑎𝑂𝐻 ×
0.5111 𝑔 𝐾𝐻𝑃 ×
𝑀 𝐻𝐶𝑙 =
39.995 𝑔 𝑁𝑎𝑂𝐻
× 0.5 𝐿 ≅ 2 𝑔 𝑁𝑎𝑂𝐻
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
1 𝑚𝑜𝑙 𝐾𝐻𝑃
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
1
×
×
= .0917𝑀 𝑁𝑎𝑂𝐻
204.22 𝑔 𝐾𝐻𝑃 1 𝑚𝑜𝑙 𝐾𝐻𝑃 . 0273 𝐿 𝑁𝑎𝑂𝐻
(𝑀 𝑁𝑎𝑂𝐻)(𝑉𝑜𝑙. 𝑁𝑎𝑂𝐻) (. 0917 𝑀)(25.0 𝑚𝐿)
=
= .0883 𝑀 𝐻𝐶𝑙
𝑉𝑜𝑙. 𝐻𝐶𝑙
27.2 𝑚𝐿
. 0897𝑀 𝐻𝐶𝑙 × .0569 𝐿 𝐻𝐶𝑙 (𝑏𝑟𝑜𝑚𝑜𝑐𝑟𝑒𝑠𝑜𝑙) = 0.0051 𝑚𝑜𝑙
. 0960 𝑀 𝑁𝑎𝑂𝐻 × .0500 𝐿 𝑁𝑎𝑂𝐻 (𝑎𝑑𝑑𝑒𝑑) = 0.0048 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 (𝑖𝑛𝑖𝑡𝑖𝑎𝑙)
0.0897 𝑀 𝐻𝐶𝑙 × .0297 𝐿 𝐻𝐶𝑙 (𝑝ℎ𝑒𝑛𝑎𝑙𝑝ℎ𝑡ℎ𝑎𝑙𝑒𝑖𝑛) ×
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
= 0.00266 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 (𝑒𝑥𝑐𝑒𝑠𝑠)
1 𝑚𝑜𝑙 𝐻𝐶𝑙
[. 0048 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 (𝑖𝑛𝑖𝑡𝑖𝑎𝑙)] − [. 00266 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 (𝑒𝑥𝑐𝑒𝑠𝑠)] = 0.00214 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 (𝑟𝑒𝑎𝑐𝑡𝑒𝑑)
0.00214 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 ×
1 𝑚𝑜𝑙 Bicarbonate
= 0.00214 𝑚𝑜𝑙 𝐵𝑖𝑐𝑎𝑟𝑏𝑜𝑛𝑎𝑡𝑒
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
[. 0051 𝑚𝑜𝑙 𝐶𝑎𝑟𝑏𝑜𝑛𝑎𝑡𝑒 & 𝐵𝑖𝑐𝑎𝑟𝑏𝑜𝑛𝑎𝑡𝑒] − [. 00214 𝑚𝑜𝑙 𝐵𝑖𝑐𝑎𝑟𝑏𝑜𝑛𝑎𝑡𝑒] = 0.00296𝑚𝑜𝑙 𝐶𝑎𝑟𝑏𝑜𝑛𝑎𝑡𝑒
. 00296 𝑚𝑜𝑙 𝐶𝑎𝑟𝑏𝑜𝑛𝑎𝑡𝑒 ×
. 00214 𝑚𝑜𝑙 𝐵𝑖𝑐𝑎𝑟𝑏𝑜𝑛𝑎𝑡𝑒 ×
60.007 𝑔 𝐶𝑎𝑟𝑏𝑜𝑛𝑎𝑡𝑒
1
×
× 100 = 7.12%
1 𝑚𝑜𝑙 𝐶𝑎𝑟𝑏𝑜𝑛𝑎𝑡𝑒
2.50 𝑔 𝑈𝑛𝑘𝑛𝑜𝑤𝑛
61.014 𝑔 𝐵𝑖𝑐𝑎𝑟𝑏𝑜𝑛𝑎𝑡𝑒
1
×
× 100 = 5.22%
1 𝑚𝑜𝑙 𝐵𝑖𝑐𝑎𝑟𝑏𝑜𝑛𝑎𝑡𝑒
2.50 𝑔 𝑈𝑛𝑘𝑛𝑜𝑤𝑛
(7.1212 − 6.2599)2 + (6.8628 − 6.2599)2 + (4.7958 − 6.2599)2
√
= 1.27453
2
Conclusions: In the 2.5 gram sample of unknown A, that my partner and I massed; .178 grams were
Carbonate and .1305 grams were Bicarbonate. Any error would be due to human error during the
titrations, and or errors in the calculations.
Primary Standards are not standardized by using other standards and are used to calculate secondary
standard. A secondary standard is a solution at has be standardized using a primary standard. A titrant is
a solution that has a known concentration, which is added to another solution to determine its
concentration. Indirect titration( also called back titration) is finding the concentration of something by
looking at how much reagent was need to titrate a second reagent, while knowing the concentration of
the first reagent.