A buffer contains 1.0mol CH3CO2H and 1.0mol CH3CO2^- diluted with water to 1.0L. How many moles of NaOH are required to increase the pH of the buffer to 5.10? (pKa oof Ch3CO2H=4.74)
You want to use the Henderson-Hasselbalch equation,
pH = pKa + log (base/acid)
First, plug in 5.10 for pH and 4.74 for pKa, and solve the H-H equation for B/A ratio.
Then look at the equation if you add OH^- to the buffer solution. It will neutralize acid to produce more acetate at the expense of CH3COOH.
CH3COOH + OH^- ==> CH3COO^- HOH
Here is what I would do.
Let x = moles OH^- to be added which is what the problem asks for.
You are starting with 1 mole CH3COOH and 1 mole CH3COO^-
At the end of x moles OH^-, the concns are
1-x for CH3COOH (acid) and 1+x for CH3COO^- (base).
Now plug those into the B/A you solved for above and solve for x.
Then to make sure you are right, plug B and A back into the H-H equation and see if you end up with pH of 5.01. Post your work if you get stuck. It may sound complicated but it really is straight forward.
To find out how many moles of NaOH are required to increase the pH of the buffer to 5.10, we first need to understand the chemistry behind it.
In this case, we have a buffer solution consisting of acetic acid (CH3CO2H) and its conjugate base, acetate ion (CH3CO2^- or CH3COO^-). The pH of a buffer is determined by the ratio of the concentrations of the acid and its conjugate base.
The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa of the acid and the concentrations of the acid and its conjugate base:
pH = pKa + log([A-]/[HA])
Where pH is the desired pH (5.10 in this case), pKa is the acid dissociation constant (given as 4.74), [A-] is the concentration of the conjugate base (CH3CO2^- or CH3COO^-), and [HA] is the concentration of the acid (CH3CO2H).
We are given that the initial concentrations of both the acid and the conjugate base are 1.0 mol each, diluted to a total volume of 1.0L. Therefore, the initial concentrations are:
[HA] = 1.0 mol/1.0 L = 1.0 M
[A-] = 1.0 mol/1.0 L = 1.0 M
Now, let's rearrange the Henderson-Hasselbalch equation for [A-]:
pH - pKa = log([A-]/[HA])
[A-]/[HA] = 10^(pH - pKa)
Substituting the values, we get:
[CH3CO2^-]/[CH3CO2H] = 10^(5.10 - 4.74)
[CH3CO2^-]/[CH3CO2H] = 10^0.36
[CH3CO2^-]/[CH3CO2H] ≈ 2.3
Since the ratio of the concentrations of the conjugate base to the acid is approximately 2.3, we can say that for every 2.3 moles of CH3CO2H, we have 1 mole of CH3CO2^-.
Now, we want to increase the pH of the buffer solution by adding NaOH, which will react with the acid and shift the equilibrium. The reaction between NaOH and CH3CO2H will result in the formation of water and the salt sodium acetate (CH3COONa):
CH3CO2H + NaOH → CH3COONa + H2O
Since the ratio of the concentrations of CH3CO2^- to CH3CO2H is 1:2.3, we need to maintain this ratio in order to keep the buffer capacity. Therefore, we need to add NaOH to convert 2.3 moles of CH3CO2H into CH3CO2^-. This means we need to add 2.3 moles of NaOH.
So, the answer is 2.3 moles of NaOH.