The Ka and Molar Mass of a Monoprotic Weak Acid

Topics: PH, Acid, Acid dissociation constant Pages: 7 (1680 words) Published: November 18, 2012
The Ka and Molar Mass of a Monoprotic Weak Acid

Chemistry Lab 152
Professor: James Giles
November 7, 2012


The purpose of this experiment was to determine the pKa, Ka, and molar mass of an unknown acid (#14). The pKa was found to be 3.88, the Ka was found to be 1.318 x 10 -4, and the molar mass was found to be 171.9 g/mol.


Acids differ considerable as to their strength. The difference between weak and strong acids can be as much as 10 orders of magnitude. Strong acids dissociate more completely than weak acids, meaning they produce higher concentrations of the conjugate base anion (A-) and the hydronium cation (H30+) in solution.

HA(aq) + H20 (( A- + H3O+

With the following formula the degree to which an acid dissociates (Ka) can be calculated and given a numerical value.

Ka = [A-][H3O+] / [HA]

Ka is the conventional way of measuring an acid’s strength. The purpose of this experiment was to determine the Ka of an unknown acid, along with its pKa and molar mass.


The unknown acid for this experiment was #14.

The experiment began with the preparation and standardization of NaOH solution. It was calculated that 2.00 grams of NaOH pellets were needed to prepare 0.5 L of 0.1 M NaOH solution. The solution was then standardized by conducting three titration trials. It was calculated that 0.7148 grams of KHP were necessary to neutralize 35 mL of the 0.1 M NaOH. Three samples of KHP were weighed approximating this number (Table 1). Each sample was mixed with 40 mL of deionized water and 2 drops of phenolphthalein in 3 Erlenmeyer flasks. Each flask was then titrated with the NaOH to a light pink endpoint. The volumes of NaOH were recorded, averaged, and the standardized. The molarity of the NaOH was found to be 0.0981.

Assuming a molar mass of 100 g/mol, it was calculated that 0.3930 g of acid was needed to neutralize 40 mL of the standardized NaOH solution. This amount was weighed out on an electronic balance to full precision and added to a clean 250 mL beaker. The acid was first diluted with 10 mL of isopropanol and then 90 mL of water. A pH meter was immersed in the acid solution and an initial pH reading of 2.61 was recorded. A buret filled with the NaOH solution was incrementally added to the acid solution and the changing pH values were recorded (Table 2). As the pH meter approached the equivalence point the amount of NaOH added each time was reduced. As the Table 2 shows, the pH rose significantly with the addition of little NaOH over this interval.

This information was plotted using Graphical Analysis producing a titration curve graph of pH vs. NaOH (Graph 1). Additional calculations and graphs were produced to help identify the equivalence point: ΔpH/ΔV vs. NaOH (Graph 2) and Vtotal x 10-ph vs. NaOH (Graph 3) Tables and Calculations

Preparation of 500 mL of 0.1 M NaOH

M = moles / volume

0.1 M NaOH = moles NaOH / 0.5 L H20 = 0.05 moles NaOH

0.05 moles NaOH x 39.986 g/mol NaOH = 1.99 g NaOH

Preparation of KHP

0.1 M NaOH = moles NaOH / 0.035 mL NaOH = .0035 moles NaOH

0.0035 moles KHP x 204.233 g/mole KHP = 0.7148 g KHP

Table 1: NaOH Titration Trials

|Trial |KHP |NaOH (to titrate to endpoint) | | |(grams) |(mL) | |1 |0.7159 |35.75 | |2 |0.7147 |35.65 | |3 |0.7149 |35.6 | | | | Avg. 35.66 |

Standardization of NaOH

0.0035 moles NaOH / .03566 mL NaOH = 0.0981 M NaOH

Table 2: pH vs. NaOH Values

|NaOH |pH |NaOH |pH |NaOH |pH |NaOH |pH | |(mL) | |(mL) | |(mL) | |(mL)...
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