Problem 5 Electrochemistry applications (20 Points) You are interested in the ac

Question

Problem 5 Electrochemistry applications (20 Points) You are interested in the acid dissociation constant for formic acid. In order to find this value, you set up the following cell. However, you want to statistically compare your value to that in the literature (1.80 x 104), so you repeat the experiment multiple times The data (including specific concentrations used) is given below the cell notation. From this data, calculate the acid dissociation constant for formic acid. Assume 1.00 bar H20g cd(s) l cdcl (aq, xM) ll Formic Acid (aq, xM), Sodium Formate (aq, xM) l H2 (g) I Pt(s) Trial THAT TA- ICdCI21 LV(measured 1 0.000404 0.00027 0.01585 0244 20.000182 0.000181 0.01301 0.236 30.000143 0.000 161 0.01526 0.231 40.000991 0.000423 0.01362 0.258 5 0.000152 0.000166 0.01635 0231 6 0.001698 0.000553 0.01414 0.264

Explanation / Answer

Oxidation half cell: Cd(s) ----> Cd2+(aq) + 2e- ; Eo = 0.40 V

Reduction half cell: 2H+(aq) + 2e- ----> H2(g) ; Eo = 0 V

Eocell = 0.40+0 = 0.40

Nernst equation for the above cell is:

Ecell = Eocell - (0.0591/2) * log ( [Cd2+]/ [H+]2 )

The [Cd2+] part is equal to [CdCl2] , and the [H+] part is obtained from the Henderson Hasselbach equation, which is written as :

pH = pKa + log ( [A-] / [HA] )

I am showing the work for trial 1, you follow the same procedure for other trials ( or simply replace the values in the methodology I am giving below ) to calculate values for other trials and take and average of those.

Trail 1:

pH = pKa + log(0.00027/0.000404) = pKa - 0.175

Thus, [H+] = Ka*100.175 = 1.49*Ka

[CdCl2] = 0.1585 ; Ecell = 0.244

Thus, 0.244 = 0.40 - (0.0591/2)* log(0.1585 / (1.49*Ka)2)

Solving we get:
Ka = 6.128*10-4

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