Please show work Calculate the pH when a 10-mL aliquot of 0.100 M alanine (H_2 A
ID: 474648 • Letter: P
Question
Please show work
Calculate the pH when a 10-mL aliquot of 0.100 M alanine (H_2 A^+) is titrated with 5.00 mL of 0.100 M NaOH. For alanine, pK_a1 = 2.35 and pK_a2 = 9.87. A 20.00 mL sample of 0.100 M NaH_2 PO_4 is titrated with 0.200 M HCl and 0.200 M NaOH. The acid dissociation constants for phosphoric acid (H_3 PO_4) are: K_a1 = 7.50 times 10^-3, K_a2 = 6.20 times 10^-8, K_a3 = 4.20 times 10^-10. a. Calculate the pH after the addition of 10.00 mL of HCl. b. Calculate the pH after the addition of 10.00 mL of NaOH. What form of glycine would you expect to predominate at a pH of 8? For glycine, K_a1 = 3.16 times 10^-3 and K_a2 = 2.51 times 10^-10.Explanation / Answer
1. Alanine (H2A+) is a weak acid and undergoes step-wise dissociation as
H2A+ (aq) -------> HA (aq) + H+ (aq); pKa1 = 2.35
HA (aq) ------> A- (aq) + H+ (aq); pKa2 = 9.87
Moles of H2A+ in the sample = (10 mL)*(1 L/1000 mL)*(0.100 mol/L) = 0.001 mol.
Moles of NaOH added = (5 mL)*(1 L/1000 mL)*(0.100 mol/L) = 0.0005 mol.
Consider the titration reaction and set up the ICE chart:
H2A+ (aq) + NaOH (aq) --------> NaHA+ (aq) + H2O (l)
initial 0.001 0.0005 0
change -0.0005 -0.0005 +0.0005
equilibrium 0.0005 0 0.0005
Volume of solution = (10 + 5) mL = 15 mL.
[H2A+]eq = (0.0005/15) mM
[NaHA+]eq = (0.0005/15) mM
Use Henderson-Hasslebach equation. Since the system will contain both H2A+ and HA, we must use pKa1. Therefore,
pH = pKa1 + log [NaHA]/[H2A+] = 2.35 + log (0.0005/15)/(0.0005/15) = 2.35 + log (1) = 2.35.
The pH of the solution is 2.35 (ans).
2. Consider the step-wise dissociations of phosphoric acid, H3PO4.
H3PO4 (aq) -------> H+ (aq) + H2PO4- (aq); Ka1 = 7.50*10-3
H2PO4- (aq) ------> H+ (aq) + HPO42- (aq); Ka2 = 6.20*10-8
HPO42- (aq) ------> H+ (aq) + PO43- (aq); Ka3 = 4.20*10-10
(a) NaH2PO4 is amphoteric and reacts with both acid and base.
Moles of NaH2PO4 present = (20.00 mL)*(1 L/1000 mL)*(0.100 mol/L) = 0.002 mol.
Moles of HCl added = (10.00 mL)*(1 L/1000 mL)*(0.200 mol/L) = 0.002 mol.
HCl completely neutralizes NaH2PO4 as below:
NaH2PO4 (aq) + HCl (aq) -------> H3PO4 (aq) + NaCl (aq).
H3PO4 will control the pH of the solution. Moles of H3PO4 formed = moles of HCl added = 0.002 mol.
Volume of the solution = (20.00 + 10.00) mL = 30.00 mL.
Initial concentration of H3PO4 formed = (0.0002 mol/30.00 mL)*(1000 mL/1 L) = 0.0667 M.
Consider the dissociation reaction as below:
H3PO4 (aq) -------> H+ (aq) + H2PO4- (aq)
initial 0.0667 0 0
change – x +x +x
equilibrium (0.0667 – x) x x
Write down the equilibrium constant as
Ka1 = [H+][H2PO4-]/[H3PO4] = (x).(x)/(0.0667 – x) = x2/(0.0667 – x)
Make an approximation; since Ka1 is small, 0.0667 >> x and hence
Ka1 = x2/0.0667
====> 7.50*10-3 = x2/0.0667
====> x2 = 5.0025*10-4
====> x = 0.0224
Therefore, [H+] = 0.0224 M and pH = -log [H+] = -log (0.0224) = 1.649 1.65 (ans).
(b) Moles of NaOH added = (10 mL)*(1 L/1000 mL)*(0.200 mol/L) = 0.002 mol.
NaOH completely neutralizes NaH2PO4 as
NaH2PO4 (aq) + NaOH (aq) -------> Na2HPO4 (aq) + H2O (l)
Moles of HPO42- formed = moles of NaOH added = 0.002 mol; volume of solution = 30.00 mL so that the initial concentration of HPO42- is 0.0667 M (check part a above).
HPO42- is formed which re-establishes equilibrium as
HPO42- (aq) --------> H+ (aq) + PO43- (aq)
Let y = [H+] = [PO43-] at equilibrium. Therefore,
Ka3 = (y).(y)/(0.0667 – y)
Use small y approximation as above and write
4.20*10-10 = y2/0.0667
===> y2 = 2.8014*10-11
===> y = 5.2928*10-6
Therefore, [H+] = 5.2928*10-6 M and pH = -log [H+] = -log (5.2928*10-6) = 5.276 5.28 (ans).