Objectives

• To determine the molar mass of an acid.
• To determine Ka for the unknown acid.

Discussion

At the half-equivalence mark, precisely one half of NaOH is needed to offset the acid completely. At this juncture, the molarity of the acid in the solution, or HA, is equivalent to the molarity of its conjugate base; this is symbolized by the equation (HA) = (A). The equation mentioned above can be derived to produce another equation, Ka = [H3O + (aq)]. By using the negative logarithm of both sides of the second equation, another equation can be derived:

– log Ka = – log [H3O + (aq)]. When simplified, this leads to the equation (pKa) = (pH); this means that at the half-equivalence mark, the pKa of the unknown acid is identical to the solution’s pH. Consequently, the acid’s Ka is derived from the value of the pKa in the following manner: Ka = 10-pKa.

Given the concentration and the volume in liters, it is possible to compute the number of moles in a solution using the equation M = n/V, where M is the concentration (molarity), n is the number of moles and V is the volume. In attempting to calculate the number of moles, this equation can be rearranged as n = M*V.The following equation represents the correlation between mass, molar mass, and moles: molar mass = mass/moles.

Experimental Procedure

The procedure is available at https://www.youtube.com/watch?v=CzcmCqVhcYc&feature=youtu.beand https://www.youtube.com/watch?v=BD_H8PzGWoc&feature=youtu.be

Data

1. Determining Molar Mass of the Unknown Acid
 Trial 1 Trial 2 Trial 3 Sample Mass 0.405 0.405 0.406 Initial reading NaOH buret 0.0ml 0.5ml 0.0ml Final reading NaOH buret 34.3ml 34.3ml 34.0ml Volume NaOH used 34.3ml 33.8ml 34.0ml Moles H+ in sample 0.00343 0.00338 0.0034 Unknown Acid Molar mass (MM) 118.08 119.82 119.41

1. Determination of Ka for the Unknown Acid
 Trial 1 Trial 2 Trial 3 NaOH volume at halfway -0.5mL pH of this solution N/A 17.0 16.5 NaOH volume at halfway pH of this solution N/A 17.5 17.0 NaOH volume at halfway +0.5mL pH of this solution N/A 18.0 17.5 Ka of acid N/A 10-6 10-6

Calculations

At the half-equivalence mark,the pH of the solution was 6.0. So far, it has already been determined that pKaequals pH which equals 6.0, which equals 10-6. Based on the equation Ka = 10-pKa, Ka, therefore, equals 10-6.

Using the equation n = M*V, the number of moles = 0.10 (as given in the experiment) * 0.0343 (34.3/1000), which is 0.00343 for trial 1, 0.00338(0.1*33.8/1000)for trial 2, and 0.0034(0.1*34.3/1000)for trial 3. The number of moles, as determined above, is 0.00343. The mass of the sample, as indicated in the experiment, was 0.405 for trial 1, 0.405 for trial 2, and 0.406 for trial 3. As a result, the molar mass is 118.08 (0.405/0.00343) for trial 1, 119.82 for trial 2 (0.405/0.00338) and 119.41 for trial 3 (0.406/0.0034).

Conclusion

In short, the molecular mass of the acid is118.08 for trial 1, 119.82 for trial 2, and 119.41 for trial 3. The Ka of the unknown acid is 10-6.

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