Example:
The problem we did in class, at least from what I wrote down, does not follow what the book says. According to Le'Chatelier's principle, adding a strong acid to a weak acid should decrease the degree of ionization for the weak acid, since adding more product (H3O+) will cause the reaction to shift to the left. For the problem we did in class, what I wrote down has the reaction shifting to the right. So instead, see the example from the book, pages 672-673.
Example:
What is the pH for a solution of 0.5 M HClO and 1.5 M HBr?
The common ion here is H+ (H3O+). Since we know HBr is a strong acid, we also know it will dissociate 100% in water, adding 1.5M H3O+ to the solution.
HClO (aq) + H2O (l) ⇌ H3O+ (aq) + ClO- (aq)
Ka = x(1.50 + x) / (0.50 – x) → approximation holds
3.5 x 10-8 = (1.50x) / (0.50)
x = 1.17 x 10-8
[H3O+] = 1.50 + 1.17 x 10-8 = 1.50 M
pH = -log(1.50) = -0.18
Note: It is possible to have a pH greater than 14 and less than 0. For example, if you have greater than 1M of a strong acid, the pH will be less than zero.
Also, the pH of a solution containing a weak acid and a strong acid will generally be equal to the pH of the strong acid alone.
Buffers:
(Basically the same as a common ion problem.)
Buffer: a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it.
Buffers contain either a weak acid and its conjugate base in a salt or a weak base and its conjugate acid in a salt. Consider a buffer that contains approximately equal concentrations of a weak acid HA and it's conjugate base A-. When a strong acid is added to the buffer, it supplies hydronium ions that react with the base A- :
H3O+ (aq) + A- (aq) → HA (aq) + H2O (l)
When a strong base is added to the buffer it supplies hydroxide ions which react with the acid HA:
OH- (aq) + HA (aq) ⇌ H2O (l) + A- (aq)
A buffer solution resists changes in pH through its ability to combine with both H3O+ and OH- ions.
Examples of buffers:
HCN and NaCN (notice the common ion CN-)
NH3 and NH4Cl
Example:
What is the pH of a solution of 0.400 M HC2H3O2 and 0.200 M NaC2H3O2?
HC2H3O2 → weak acid → HA
NaC2H3O2 → salt containing conjugate base → C2H3O2- → A-
Note: Na+ doesn't do much but show up. Na+ is like the person C2H3O2- brought to the party then didn't hang out with once they got there. Which is just mean. So C2H3O2- is kind of an asshole (A-). Maybe that's an easy way to remember it.
But we know 0.200 M NaC2H3O2 will dissociate to yield 0.200 M C2H3O2-.
HC2H3O2 (aq) + H2O (l) ⇌ H3O+ (aq) + C2H3O2- (aq)
Notice that there is a nonzero initial concentration of A-. This is from the salt.
Ka = x(0.200 + x) / (0.400 – x) → approximation holds
1.7 x 10-5 = x(0.200) / (0.400)
x = 3.4 x 10-5 = [H3O+]
pH = -log(3.4 x 10-5) = 4.47
Example:
What is the pH of a solution that is 0.45 M HF and 0.115 M NaF?
HF (aq) + H2O (l) ⇌ H3O+ (aq) + F- (aq)
Ka = x(0.115 + x) / (0.45 – x) → approximation holds
6.8 x 10-4 = x(0.115) / (0.45)
x = 0.00266 = [H3O+]
pH = -log(0.00266) = 2.57
Example:
For a buffer containing HC2H3O2 and NaC2H3O2, which will react with NaOH?
Consider NaOH to be OH-.
HC2H3O2 + OH- → H2O + C2H3O2-
Which will react with HCl?
Consider HCl to be H+ (H3O+).
C2H3O2- + H3O+ → HC2H3O2 + H2O
or A- + H3O+ → HA + H2O
Note: When acids and bases are added to a buffer, the pH will change, but only slightly. The pH does not stay exactly the same.
Example:
THIS IS THAT HUGE 3 PART PROBLEM REGARDING BUFFERS
(Note: Ka for HC3H5O2 = 1.3 x 10-5)
What is the pH for a buffer made from MIXING 47.0 mL of 0.678 M HC3H5O2 and 33.0 mL of 0.533 M NaC3H5O2?
Note: In this problem, we are making a buffer.
HC3H5O2 → weak acid → HA
NaC3H5O2 → salt with conjugate base → NaA → A-
Whenever you see the words MIXING or TITRATING it is a clue that the volume is going to change, so molarity will change. The first step in these problems is always to convert to moles:
In order to do that we need to convert mL to L:
(47.0 mL) x (1 L / 1000mL) = 0.047L
(33.0 mL) x (1L / 1000mL) = 0.033L
(0.047L) x (0.678 mols / L HA) = 0.0319 mols HA
(0.033L) x (0.533 mols / L A-) = 0.0176 mols A-
Now divide by the new volume to get the new molarity:
New volume = 0.047L + 0.033 L = 0.080 L
0.0319 mols HA / 0.080 L = 0.399 M HA
0.0176 mols A- / 0.080 L = 0.220 M A-
HA + H2O ⇌ H3O+ + A-
Ka = x(0.220 + x) / (0.399 – x) → approximation holds
1.3 x 10-5 = x(0.220) / (0.399)
x = 2.36 x 10-5 = [H3O+]
pH = -log(2.63 x 10-5) = 4.63
Now find the pH when 95.0 mL of the buffer we've just created is mixed with 5.00 mL of 0.100 M HCl.
We know that it is the conjugate base in our buffer that will react with the HCl (the A-).
We also know that HCl is a strong acid that dissociates 100 % in water.
When mixing, convert to moles:
(0.399 M HA / L)(0.095 L) = 0.379 mols HA
(0.220 M A- / L)(0.095 L) = 0.0209 mols A-
(0.100 M HCl / L)(0.005 L) = 5 x 10-4 mols HCl
Now we need to look at how the conjugate base of the buffer and the HCl are going to react and how much HA they are going to produce:
HCl + A- → HA + Cl-
We subtract the mols of HCl from the left side of the reaction because the HCl will be used up before the A- is used up, there is less of it (limiting). And since this is what we are subtracting from the left side, it is also what we are adding to the right side. So now:
A- = 0.0204 mols
HA = 0.0384 mols
Remember we're still in mols, so we need to divide by our new total volume to get molarity:
A- = 0.0204 mols / 0.1 L = 0.204 M
HA = 0.0384 mols / 0.1 L = 0.384 M
0.384 M is our new acid concentration. What happens when it dissociates in water? We get our final hydronium ion concentration from which we can calculate pH:
HA + H2O ⇌ H3O+ + A-
Ka = x(0.204 + x) / (0.384 – x) → approximation holds
1.3 x 10-5 = x(2.04) = (0.384)
x = 2.45 x 10-5 = [H3O+]
pH = -log(2.45 x 10-5) = 4.61
Now find the pH when 95.0 mL of the buffer we initially created is mixed with 5.00 mL of 0.350 M NaOH:
Since we're dealing with a changing volume, first thing to do is convert to moles:
(0.399 mols / 1 L HA)(0.0950L) = 0.0379 mols HA
(0.220 mols/ 1 L A-)(0.0950L) = 0.0209 mols A-
(0.350 mols/ 1 L NaOH)(0.005L) = 0.00175 mols OH-
Which component of the buffer will the NaOH react with? The HA.
OH- + HA → H2O + A-
We know to subtract 0.00175 mols because there is less of it and everything is 1:1. Now take the new values for HA and A-, which are in mols, and divide them by the new volume to get the concentration in molars:
HA = 0.0362 mols / 0.1 L = 0.362 M HA
A- = 0.0227 mols / 0.1 L = 0.227 M A-
Now this acid will dissociate in water to yield H3O+, from which we can find the new pH:
HA (aq) + H2O (l) ⇌ A- (aq) + H3O+ (aq)
Ka = x(0.227 + x) / (0.362 – x) → approximation holds
1.3 x 10-5 = x(0.227) / (0.362)
x = 2.07 x 10-5 = [H3O+]
pH = -log(2.07 x 10-5) = 4.68
Notice that the pH of the original buffer was 4.63. When we added a strong acid, the pH only dropped to 4.61, and when we added a strong base, the pH went up to 4.68. So although the pH didn't stay exactly the same, it stayed pretty stable.
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