Equilibrium constant, Ka By Soo Jeon

background Most weak acids dissociate 5% or less, that is, 95% or more of the acid remains as HA. The smaller the value of Ka, the weaker the acid.

Background.. Monoprotic acids are those acids that are able to donate one proton per molecule during the process of dissociation (sometimes called ionization) as shown below (symbolized by HA):protondissociation HA(aq) + H 2 O(l) -> H 3 O + (aq) + A - (aq)

scenario A student prepared three solutions by adding 12.00, 18.5, and 35.00ml of 7.5x10 -2 M NaOH solution to 50.00ml of a 0.100M solution of a weak monoprotic acid, HA. The solutions were labeled X,Y,and Z respectively. Each of the solutions were diluted to a total volume of ml with distilled water. The pH readings of these solutions, obtained through the use of a calibrated pH meter, were X,Y, Z. We are going to determine Ka for a monoprotic weak acid.

Purpose This experiment will determine Ka for a monoprotic weak acid using pH value and A-ion concentration.

Materials 7.5x10 -2 M NaOH solution 0.100M solution of a weak monoprotic acid distilled water calibrated pH meter 100 ml of Three beakers

pH of each solution Using the calibrated pH meter, find pH of each solution. X : 6.50 Y : 6.70 Z : 7.10

Calculations 1.Convert pH to equivalent H30+ concentration. 1) pH = -log [H + ] [H ] = 10 -pH X : = 3.16 x M H Y : = 2.00 x M H Z : = 7.94 x M H 3 0 +

2. Calculate the number of moles of acid added The moles of acid was constant for each solution ml x 1L x moles HA ml 1 L = 5.00 x mol HA

3. Calculate the number of moles of OH - added X : 12.00ml x 1L x 7.50 x mol OH ml 1 L = 9.00 x mol OH- Y : 18.50ml x 1L x 7.50 x mol OH ml 1L = 1.39 x mol OH- Z : 35.00ml x 1L x 7.50 x mol OH ml 1L = 2.69 x mol OH-

4. Determine the concentration of HA and A - ion X : [HA] = initial moles HA – moles OH - added volume of solution = 5.00 x x L = 4.10 x Molarity Moles A - formed = moles OH - added [ A-] = 9.00 x mol A - = 9.00 x Molarity 0.100L

Y : [HA] = 5.00 x – 1.39 x L = 3.61 x Molarity [A-] = 1.39 x mol A L = 1.39 x Molarity

Z : [HA] = 5.00 x x L = 2.37 x Molarity [A - ] = 2.63 x mol A L = 2.63 x Molarity

5. Reciprocal of A - ion concentration To show that the concentration is increasing, we need to flip the A - ion concentration. X : 1/(9.00 x 10 -3) = 1.11 x M -1 Y : 1/(1.39 x ) = 7.19 x 10 1 M -1 Z : 1/(2.63 x ) = 3.80 x 10 1 M -1

6. Reciprocal of A - ion concentration against H 3 O + ion concentration If it is not reciprocal of A- ion concentration, then the slope (K a ) would be negative value.

7. Determine the slope of the line Slope = ΔY = Δ[1/A - ] ΔX = Δ[H ] =1.11 x 10 2 – 3.80 x 10 1 = 3.08 x 10 8 (3.16 x ) – (7.94 x )

8. Determine the initial concentration of the acid, HA. = 5.00 x mol HA = 5.00 x M L

9. Determine the Ka for the weak acid. 1 = Δ[H ] = (H ) · (A - ) Slope Δ[1/A - ] Ka = 1 = 1___________ HA x slope (5.00 x )·(3.08 x 10 1 ) = 6.49 x M

Conclusion In this lab, I measured pH values of each solution and then determined the concentrations of HA and A- ion. I calculated the slope of the reciprocal of A - ion concentration on the ordinate against the H 3 O + ion concentration and got the Ka for the weak acid, 6.49 x M.

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