Explanation below. Here's the TL;DR:
Ka of Acetic Acid = 1.8 E -5 (from reference Google);
pKa = 4.74 for HA (Acetic Acid).
pH = pKa + log(A-/HA)
4.6 = 4.74 +log(ratio)
0.72 = ratio
x moles of HNO3, a strong acid, will be subtracted from conj base A- (0.1 moles - x) and added to weak acid HA (0.01 moles + x), to make a ratio of 0.72:
A-/HA = 0.72 = (0.1 - x)/ (0.01 + x)
x = 0.054 moles of HNO3.
Use x to find new values:
A- = (0.1 moles - x); HA = (0.01 moles + x)
A- = 0.046 moles; HA = 0.064 moles.
Check:
pH = 4.74 + log(0.046/0.064) = 4.599 = 4.6
x = 0.054 moles of 10.0 M HNO3 = 5.4 mL.
Here's the detailed explanation:
To calculate the pH of an aqueous buffer solution, you can use the Henderson-Hasselbach equation.
1) pH = pKa + log( [A-] / [HA]) to calculate the pH of a buffer of a weak acid and conjugate base.
In this case, our weak acid HA is Acetic Acid, and our conjugate base A- is Sodium Acetate.
pH calculations can get really long, so it's a good idea to look for shortcuts: for example, in this case, we can see that the concentration of the conjugate base (0.100 M Sodium Acetate) is 10x stronger than the concentration of the weak acid (0.0100 M Acetic Acid). That means that
2) pH = pKa + 1 if there is 10x more conjugate base than weak acid
assuming we don't add anything else to the buffer.
So, maybe we don't need to add any of that strong Nitric Acid (10.M HNO3) to the buffer. Let's check and see if we already have the correct pH with no added strong acid.
So, we're checking to see if pKa + 1, which is our current buffer's pH, is equal to the required 4.6.
To find the pKa, we need the Ka of the weak acid. Google indicates the Ka of Acetic Acid = 1.8 x 10-5.
3) pKa = -logKa to calculate pKa from Ka
pKa = -log(1.8 x 10-5) = 4.74.
So, our current buffer pH is pKa + 1, or 5.74. (You can verify this using equation 1, the Henderson-Hasselbach equation, above). As expected, it's not the pH of 4.60 that the question requires, unfortunately. That means we need to add some of that strong Nitric Acid to our buffer to lower the pH from its current 5.74, to the required 4.6.
At this point, we need to calculate how many moles of Acetic Acid and Sodium Acetate are needed to achieve a pH of 4.60. Henderson-Hasselbach can get us started with that:
From equation 1: pH = pKa + log( [Sodium Acetate]/[Acetic Acid]).
It looks like the best we can do is get the ratio of these moles, instead of the exact amounts of moles.
4.6 = 4.74 + log(A-/HA)
-0.14 = log(A-/HA)
10-0.14 = 10log(A-/HA) here we take the anti-log of both sides, which is 10x button in most calculators:
0.72 = A-/HA
Does it make sense that our final buffer should be a ratio of 0.72 moles of base / for every 1.0 moles weak acid?
Yes, because if it were a 1:1 ratio, our pH would be equal to the pKa (another shortcut), which is 4.74. So, we need a ratio that slightly favors the acid to bring the pH all the way down from 4.74 to the required 4.60.
Ok, we need one more equation to figure out how to get from our current 10:1 ratio, to a 0.72:1 ratio.
We're going to use a BCA table, ("Before Change After"). It's like an ICE table, but it uses moles instead of molarity. A BCA table lets us know how moles of added strong acids like our 10.0M HNO3 affect the moles of our buffer chemicals.
In this table, A- represents our conjugate base, Sodium Acetate, and HA represents Acetic Acid.
When all is said and done, we need the values in the last row of our table to make the ratio of 0.72.
So, we're going to use our knowledge of the behavior of strong acids to do a math trick: we know that however many moles of HNO3 we add, we're going to subtract that number of moles from A-, and add to HA. This is illustrated in the BCA table.We don't know how many moles of HNO3 to add yet. We're going to use x moles of HNO3. We enter the starting amounts of our Acetic Acid (0.01 moles) and Sodium Acetate (0.1 moles) in the Before row, too. These are the moles because our volume is 1 liter.
HNO3 + A- -------> HA + NO3-
Before x 0.1 0.01 0
Change -x -x +x +x
After 0 0.1 - x 0.01 + x x
Thanks to the last row of our BCA table, we can figure out how to change our 10:1 ratio, to the desired 0.72:1 ratio of A-/HA:
0.72 = A-/HA = (0.1 - x) / (0.01 + x).
Algebra reveals x = 0.054 moles of HNO3.
Let's see if it works:
HNO3 + A- -------> HA + NO3-
Before 0.054 0.1 0.01 0
Change -0.054 -0.054 +0.054 +0.054
After 0 0.046 0.064 0.054
Did we get the required ratio of A-/HA = 0.72?
0.046 / 0.064 = 0.72. Now let's check if we get the pH of 4.60 required for the problem:
pH = pKa + log ([Sodium Acetate] / [Acetic Acid])
pH = 4.74 + log (0.046 / 0.064)
pH = 4.596 = 4.6
Now we can use our value of x to answer the question: how many mL of 10.0 M HNO3 to add to our buffer to achieve this pH of 4.6. To do this, we multiple the moles, x, by the inverse molarity, 1/M = L/moles:
x = 0.054 moles * (1000 mL / 10 moles) = 5.4 mL. That's it!
