Problem 17.17 Part A Calculate the percent ionization of 7.010 ?3 M butanoic aci
ID: 791921 • Letter: P
Question
Problem 17.17 Part A Calculate the percent ionization of 7.010?3M butanoic acid (Ka=1.510?5). Express your answer using two significant figures.Part B Calculate the percent ionization of 7.010?3M butanoic acid in a solution containing 7.510?2M sodium butanoate. Express your answer using two significant figures.
Problem 17.17 Problem 17.17 Part A Calculate the percent ionization of 7.010?3M butanoic acid (Ka=1.510?5). Express your answer using two significant figures.
Part B Calculate the percent ionization of 7.010?3M butanoic acid in a solution containing 7.510?2M sodium butanoate. Express your answer using two significant figures.
Part B Calculate the percent ionization of 7.010?3M butanoic acid in a solution containing 7.510?2M sodium butanoate. Express your answer using two significant figures.
Explanation / Answer
Part A. Calculate the percent ionization of 7.0*10^-3 M butanoic acid (Ka=1.5*10^-5).
By definition, in equilibrium:
Ka = [H+][A-]/[HA]
Then
Assume that [H+] = [A-] = x … due to stoichiometry (i.e. 1 mol of H+ per each mol of A-)
Then, in equilibrium [HA] = M – x (i.e. the original concentration minus the acid in equilibrium)
Substitute
Ka = (x*x)/(M-x)
1.5*10^-5 = x*x/(7.0*10^-3-x)
solve for x (quadratic equation)
x = [H+] = 3.166*10^-4 M
[H+] = [A-] = x = 3.166*10^-4
The ionization:
% ionization = [H+] / M * 100% = (3.166*10^-4)/(7.0*10^-3) * 100% = 4.522 %
Part B. Calculate the percent ionization of 7.0*10^-3 M butanoic acid in a solution containing 7.5*10^-2 M sodium butanoate.
This is a buffer
pH= pKa + log (sodium butanoate / butanoic acid)
pKa = -log(KA) = -log(1.5*10^-5) = 4.82390
substitute values
pH= 4.82390 + log ( (7.5*10^-2 ) / (7.0*10^-3))
pH = 5.85386
now,
[H+] = 10^-ph = 10^-5.85386 = 0.00000140
now, substitute
% ionization = [H+] / M * 100% = (0.00000140)/(7.0*10^-3) * 100% = 0.02 %
that is; the acid is mostly molecular