Sally Burke Chem 1A

Section 0288 Locker: A505

Instructor: O. Raola Due date: 10/13/15

Experiment 12

Determination of the Molar Mass of an Unknown Diprotic Acid

Purpose The purpose of this lab was to determine the molar mass of an unknown diprotic acid by titrating it with a standardized solution of NaOH. Procedure First, standardize a diluted solution of Sodium Hydroxide (NaOH) with Potassium Hydrogen Phthalate (KHP) in order to find the exact molarity of the basic solution. Then titrate a solution of unknown diprotic acid with the standardized solution of NaOH in order to find its molar mass. Safety Precautions NaOH of such a high concentration (2 mol ∙ L!!) is highly corrosive, meaning that it can easily burn through living tissue. It is important to wear safety glasses while conducting this experiment. If the concentrated solution comes in contact with skin, wash thoroughly. Data Part One

Trial 1 Trial 2 Trial 3 Mass of KHP (g) 0.7107 0.7238 0.7081

Moles of KHP (mol) 3.480 × 10-3 3.544 × 10-3 3.467 × 10-3

Initial Volume (mL) 0.00 0.05 10.00 Final Volume (mL) 31.81 32.65 41.75 Volume Used (mL) 31.81 32.60 31.75

Molarity (g/mol) 0.1094 0.1086 0.1093 Average Molarity (g/mol) 0.1091

%RSD 0.3995% Part Two

Trial 1 Trial 2 Trial 3 Unkown Code 5 5 5

Mass of Unknown (g) 0.3009 0.2914 0.3062 Initial Volume NaOH (mL) 8.00 & 28.50 0.03 0.00 Final Volume NaOH (mL) 50.00 &33.55 47.51 48.34 Volume NaOH Used (mL) 47.05 47.48 48.34

Moles NaOH (mol) 5.133 × 10-3 5.180 × 10-3 5.274 × 10-3

Moles Unknown (mol) 2.567 × 10-3 2.590 × 10-3 2.637 × 10-3

Molar Mass Unknown (g/mol) 117.2 112.5 116.1 Average Molar Mass (g/mol) 115.3

%RSD 2.132% Actual Molar Mass (g/mol) 116.1 maleic acid

% Error 0.689%

Sample Calculations Part One

Moles of KHP titrated: = 𝑀𝑎𝑠𝑠 𝑜𝑓 𝐾𝐻𝑃 × 𝑀𝑜𝑙𝑎𝑟 𝑀𝑎𝑠𝑠 𝑜𝑓 𝐾𝐻𝑃

= 0.7107 𝑔 𝐾𝐻𝑃 × 1 𝑚𝑜𝑙 𝐾𝐻𝑃

204.22 𝑔 𝐾𝐻𝑃

= 3.480 × 10!! mol KHP

Volume NaOH used: Vused = Vfinal – Vinitial Vused = 41.75 mL – 10.00 mL Vused = 31.75 mL

Molarity of NaOH: *Note that the amount (moles) of KHP and NaOH are equivalent because they react in a one-to-one ratio

Molarity = AmountVolume

Molarity = 3.480 × 10!! mol NaOH

31.81 mL × 1 𝐿1000 𝑚𝐿

M = 0.1904 mol ∙ L!! NaOH

Average Molarity:

𝑋 = 𝑋𝑁

𝑋 =0.1094+ 0.1086+ 0.1093

3 = 0.1091 mol ∙ L!! %RSD:

=

𝑇𝑟𝑖𝑎𝑙 1− 𝑋 ! + (𝑇𝑟𝑖𝑎𝑙 2− 𝑋)! + (𝑇𝑟𝑎𝑖𝑙 3− 𝑋)!# 𝑜𝑓 𝑇𝑟𝑖𝑎𝑙𝑠 − 1

𝑋×100

=

0.1094− 0.1091 ! + (0.1086− 0.1091)! + (0.1093− 0.1091)!3− 1

0.1091 ×100 =0.3995%

Part Two

Volume NaOH used: See calculation in Part One

Moles of NaOH used:

Molarity = AmountVolume

Amount = Molarity × Volume

𝐴𝑚𝑜𝑢𝑛𝑡 = 0.1091 mol ∙ L!! × 47.48 𝑚𝐿 ×1 𝐿

1000 𝑚𝐿

Amount = 5.180 × 10-3 mol NaOH

Moles of Unknown titrated:

𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑁𝑎𝑂𝐻×1 𝑚𝑜𝑙 𝑈𝑛𝑘𝑛𝑜𝑤𝑛2 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻

5.180 × 10!! mol NaOH×1 𝑚𝑜𝑙 𝑈𝑛𝑘𝑛𝑜𝑤𝑛2 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻

= 2.590 × 10-3 mol Unknown

Molar Mass of Unknown:

𝑀𝑜𝑙𝑎𝑟 𝑀𝑎𝑠𝑠 = 𝑀𝑎𝑠𝑠𝐴𝑚𝑜𝑢𝑛𝑡

𝑀𝑜𝑙𝑎𝑟 𝑀𝑎𝑠𝑠 = 0.2914 𝑔 𝑈𝑛𝑘𝑜𝑤𝑛

2.590 × 10!! mol Unknown Molar Mass = 112.5 g/mol

Average Molar Mass:

𝑋 = 𝑋𝑁

𝑋 =117.2+ 112.5+ 116.1

3

𝑋 = 115.3 g/mol

%RSD:

=

𝑇𝑟𝑖𝑎𝑙 1− 𝑋 ! + (𝑇𝑟𝑖𝑎𝑙 2− 𝑋)! + (𝑇𝑟𝑎𝑖𝑙 3− 𝑋)!# 𝑜𝑓 𝑇𝑟𝑖𝑎𝑙𝑠 − 1

𝑋×100

=

117.2− 115.3 ! + (112.5− 115.3 )! + (116.1− 115.3 )!3− 1115.3 ×100

= 2.132%

% Error:

=𝐴𝑐𝑡𝑢𝑎𝑙 − 𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙

𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 ×100

=115.3− 116.1

116.1 ×100 = 0.689%

Discussion The molar mass of the unknown diprotic acid was experimentally determined to be 115.3 g ∙mol!!, after titrating it with a standardized solution of NaOH. Of the four possible acids (tartaric acid, maleic acid, malonic acid, and succinic acid), this value most closely resembles maleic acid, which has a molar mass of 116.1 g ∙mol!!. The percent error of this experiment was 0.689%, thus the results were quite accurate. The main source of systematic error would have resulted from leaving the cap off of the Nalgene bottle. This would leave the NaOH to absorb water, resulting in the calculated concentration being greater than the actual concentration. However, most error in this experiment was more likely to be random. This assumption can be made by looking at the results of the three trials. Trial 3 determined the molar mass to be 116.1 g ∙mol!!, which is equivalent to the theoretical value. Trial 1 found it to be 117.2 g ∙mol!!, which is greater than expected. Lastly, Trial 2 determined it to be 112.5 g ∙mol!!, which is smaller than the theoretical value. Systematic error is defined as causing results that are consistently greater or consistently smaller than the expected result. It is clear that this was not the case in these trials. Instead, random error played a bigger part, as evidenced by the percent relative standard deviation (%RSD) of 2.132%. One source may be not transferring all of a substance from one container to the next. Using the wrong amount of indicator can shift the observed end point, thus changing the result of that trial. The number of significant figures of the equipment used also limits precision. The determination of the molar mass of the unknown acid by titration would not have been possible without first standardizing the NaOH solution. Standardization establishes the exact concentration of the base. To standardize a solution, it is necessary to titrate it with a known acid. Interestingly, the concentration of the base could just as easily be calculated:

𝑀!𝑉! = 𝑀!𝑉!

2.0 𝑚𝑜𝑙 ∙ 𝐿!! 20 𝑚𝐿 ×1 𝐿

1000 𝑚𝐿 = 𝑀! 20 𝑚𝐿 + 375 𝑚𝐿 ×1 𝐿

1000 𝑚𝐿

𝑀! = 0.1013 𝑚𝑜𝑙 ∙ 𝐿!! The experimentally standardized value of 0.1091 𝑚𝑜𝑙 ∙ 𝐿!! came within 7.7% error of this calculated value. But that is not to say that the calculated value is completely accurate in itself. The calculation assumes that the original solution of NaOH was precisely 2.0 M, which is not a safe assumption. Plus, it also requires that the volume measurements were made accurately during experimentation. Thus, the process of standardization could find a more accurate value of molarity than the calculation. Indicators play a key role in titrations. The definition of a titration is the neutralization of an acid and a base, essentially creating water. There would be no way to see the equivalence point between clear solutions without an indicator. In the case of this experiment, a one-color indicator called phenolphthalein was used. It is clear when in acidic solution and turns pink when in a basic one:

H! + Ind! ↔ HInd Colorless ↔ Pink

However, it is important to note that indicators are not perfect. They cause something commonly known as titration error. The indicator does not change color at the equivalence point of the reaction. Instead, it indicates the end point – the point at which the solution becomes basic. Thus, the end point is actually further than the equivalence point. This means that the solution becomes neutral and then more base is added until it becomes basic. As a result, all titrations have inbuilt systematic titration error. This lab also required a basic understanding of stoichiometry. The ability to convert from moles of one compound to moles of another was key. Interestingly, this mole-to-mole conversion also played a role in the procedure of the experiment. The standardization reaction was a one-to-one ratio, so approximately 0.7 g of KHP and 20 mL DI water were used in creating the acidic solution. However, the second part was to find the molar mass of a diprotic acid. Thus, the reaction had a one-to-two ratio, and approximately 0.3 g of unknown acid and 40 mL of DI water were used. Essentially, it was made half as concentrated as the first one. If it were not, it would have required twice as much NaOH to titrate it. Based on this pattern, I predict that if a triprotic acid, such as phosphoric acid, the mass would need to be a third of the original value and the volume of DI water added would be triple. It would be interesting to conduct this experiment with a triprotic acid to test this prediction. Conclusion The reported experimentally determined molar mass was the main purpose of this experiment. I also learned how to conduct an accurate titration and learned the importance of standardization.