Determining the Ka of a Weak Acid by pH Measurements Manatee Community College
Introduction
The Bronsted-Lowry definition of an acid is a substance that donates a hydrogen ion to a Bronsted-Lowry base, which is a hydrogen ion acceptor. A typical acid-base reaction is that of hydrobromic acid (HBr) with water, as seen in Equation 1.
+ - (1) HBr(aq) + H2O(l) ⇄ H3O (aq) + Br (aq)
In this reaction, HBr, the acid, donates a hydrogen ion to a water molecule, the base. The products of this reaction are a hydronium ion, the conjugate acid, and a bromine ion, the conjugate base. Another acid-base reaction is shown in Equation 2.
+ - (2) HF(aq) + H2O(l) ⇄ H3O (aq) + F (aq)
In this reaction, a hydrofluoric acid molecule becomes a fluorine ion when it donates a + proton to a water molecule, changing the water into a H3O ion. The two reactions above seem to be very similar, when in fact they are not. The main difference between them is the strength of their acids. Hydrobromic acid, HBr, is a strong acid, whereas hydrofluoric acid, HF, is a weak acid. What this means is that Hydrobromic acid will essentially dissociate completely (100%) in water and the hydrofluoric acid dissociates only approximately 1%. The relative strength of acids is based on how well, or to what extent they dissociate in water. Strong acids are strong electrolytes, which dissociate essentially to 100% in water. Examples of strong acids are HClO4, HNO3, H2SO4, HCl, HBr and HI. Weak acids such as HF and CH3CO2H are weak electrolytes that dissociate to a much lesser extent. The extent of dissociation of any weak acid is indicated by the magnitude of its acid dissociation constant, Ka. The Ka is an equilibrium expression with respect to -3 -3 an acid and if the Ka> 1 x 10 it is considered a strong acid and if the Ka < 1 x 10 it is considered to be a weak acid. The general rule is that as the Ka increases the strength of the acid also increases. The Ka expression for Equation 2 can be seen in Equation 3.
- + (3) Ka = [F ] [H3O ] / [HF]
The Ka expression is the products divided by the reactants and the coefficients become the exponents. The brackets [ ], indicate the molar concentration (mol / L) of the species in the brackets. When we write equilibrium expressions we leave out pure liquids and solids because their concentrations remain essentially constant. There are no units for equilibrium constants such as Ka, but they are temperature dependent so we must include -5 the temperature. For example the Ka of CH3CO2H is 1.7 x 10 @ 25C. The acid dissociation constants for some weak acids @ 25C are listed in Table 1.
Ka 1 Table 1: Acid Dissociation Constants @ 25C
Acid Ka -5 Acetic acid, CH3CO2H 1.7 x 10 -10 Boric acid, H3BO3 5.8 x 10 -4 Nitrous acid, HNO2 4.5 x 10 Hypochlorous acid, HOCl 3.5 x 10-8 Hydrocyanic acid, HCN 4.0 x 10-10 Hydrofluoric acid, HF 7.2 x 10-4 Potassium Hydrogen Phthalate (KHP) 4.0 x 10-6
Calculating The Acid Dissociation Constant From pH Data
We will use the generic formula HAn to stand for a weak acid and this allows us to represent its dissociation using Equation 4.
+ - (4) HAn(aq) + H2O(l) ⇄ H3O (aq) + An (aq)
The Ka expression for Equation 4 can be seen in Equation 5.
+ - (5) Ka = [H3O ] [An ]/[HAn]
+ - In this equation [H3O ]and [An ] are identical. We can calculate the Ka for HAn if we know the concentrations of the dissociated and undissociated species in Equation 5. + We can determine the H3O concentration by pH measurements. Equation 6 is the + equation to find for pH and Equation 7 is to find for [H3O ].
+ (6) pH = -log [H3O ]
+ -pH (7) [H3O ] = 10
- + According to Equation 4 the An concentration is equal to the H3O concentration at equilibrium. If we know the original concentration of the acid, HAn, we can calculate the + undissociated HAn equilibrium concentration by difference. Once we know the [H3O ], - [An ], and [HAn] we can calculate the Ka using Equation 5.
Calculating The Acid Dissociation Constant From Neutralization Data
A second which method we will use for determining the Ka involves using pH measurements made following the partial neutralization of the weak acid with hydroxide ion (OH-). A generic equation for this reaction is shown in Equation 8.
- - (8) HAn(aq) + OH (aq) ⇄ H2O(l) + An (aq) + - A solution containing H3O , An , and HAn will be prepared by partially neutralizing the weak acid with sodium hydroxide. Due to the neutralization the HAn, + - H3O , and An concentrations are no longer equal. Instead we have created a buffer
Ka 2 solution with appreciable concentrations of HAn and An-. Because hydroxide is a strong base the reaction of OH- ions with HAn goes essentially to completion and the number of moles of An- ions formed is equal to the number of moles of NaOH used We can calculate the HAn and An- ion equilibrium concentrations if we know the volume and concentration of HAn initially and the volume and concentration of NaOH solution added. Here are the steps required to calculate HAn and An-: 1. Calculate the initial number of moles of acid: M x V. This applies for all 4 runs where M is the molarity (moles/L) and V is the volume in Liters. 2. Calculate the number of moles of OH- added for each run: M x V 3. Calculate the number of moles of HAn left after hydroxide is added by subtracting moles OH- from moles HAn. See equation 9 below. 4. The number of moles of An- is the same as the number of moles of OH- added. 5. Calculate the Molarity of HAn by dividing the moles of HAn by the total volume of liquid. 6. Calculate the Molarity of An- by dividing the moles of An- by the total volume of liquid. - + 7. Use equation 5 to find Ka. [HAn] is found in step 5. [An ] is in step 6 and [H3O ] is found from the measured pH.
(8) Equilibrium moles HAn = (initial moles of HAn) – (initial moles of OH-)
Eq. 9: Equilibrium moles HAn = (initial moles of HAn) – (initial moles of OH-)
Procedure 1. Obtain 180 mL of a 0.250 M unknown weak acid solution. Record the number of your unknown and the exact molarity of the solution. 2. Fill a clean 50 mL buret with the weak acid solution and label acid. 3. Obtain 140 mL of 0.125 M NaOH solution. Record the exact molarity. 4. Fill a clean 50 mL buret with the NaOH solution and label base. 5. Prepare solution 1 by adding 40.00 mL of the unknown acid solution from the buret to a clean 250 mL volumetric flask. Record your initial reading and final reading of the buret to the nearest 0.01 mL. a) Add 12.50 mL of the NaOH solution from the second buret to the acid solution in the volumetric flask. Record your initial and final buret reading to the nearest 0.01 mL. b) Dilute the solution to the 250 mL mark with DI water and thoroughly mix. 6. Pour about 50 mL of solution 1 into a clean 100 mL beaker and take the pH reading using the pH probe. 7. Use the amounts specified in Table 2 to prepare solutions 2-4 just as you did solution 1 and take their pH measurements. Table 2: Solution # Volume Weak Acid, mL Volume NaOH, mL 1 40.00 12.50 2 40.00 25.00 3 40.00 37.50
Ka 3 4 40.00 50.00
Data
Unknown number
Molarity of unknown acid
Molarity of NaOH
1 2 3 4
Unknown Acid Solution
Initial Buret reading, mL
Final Buret Reading, mL
Volume of unknown acid, mL
NaOH Solution
Initial Buret reading, mL
Final buret reading, mL
Volume of NaOH solution, mL Total Volume 250 250 250 250 of Solution, mL PH
Ka 4 Calculating Ka of your unknown acid
1 2 3 4
Initial number of moles of Han and NaOH before mixing
HAn(aq)
- OH (aq)
Number of moles at equilibrium after mixing
HAn(aq)
- An (aq)
Equilibrium concentrations, mol/L
HAn(aq)
- An (aq)
+ H3O (aq)
Ka
Average Ka
Which of the weak acids in Table 1 do you think your unknown acid is?
Ka 5 Pre-Lab Question
1. A student following the procedure in this lab prepared 4 solutions by adding 12.25, 23.00, 25.50, and 34.50 mL of a 8.25 x 10-2 M NaOH solution to 50.00 mL of a 0.152 M solution of weak acid. The solutions were labeled 1, 2, 3, and 4 respectively. Each of the solutions was diluted to a total volume of 250 mL with DI water. The pH readings of these solutions were: (1) 6.64 (2) 6.98 (3) 7.04 (4) 7.23 PRE-LAB DATA SHEET 1 2 3 4
Initial number of moles
HAn(aq)
- OH (aq)
Number of moles at equilibrium
HAn(aq)
- An (aq)
Equilibrium concentrations, mol/L
HAn(aq)
- An (aq)
+ H3O (aq)
Ka
Average Ka
Which of the weak acids in Table 1 do you think your unknown acid is?
Ka 6