What is the pH of a solution prepared by adding 4.000 grams of NaOH to a 50.00 mL of a buffer that is 2.00 molar both in acetic acid and sodium acetate? (Assume final volume is still 50.00 mL)
Use the Henderson-Hasselbalch equation. The addition of NaOH will decrease the acid by that many moles and increase the acetate by that many moles. Substitute into the HH equation and solve for pH. Post your work if you get stuck.
i did .1 mols of AC- + .1 mols of OH gives me .2 mols of AC.
the .2 mols of AC reacts with water
AC- + H20---> OH- + HAC
so the M is .2/.05 L = 4M
to solve for the OH conc i did X^2 / 4 = 5.6 x 10^-10. and solving for the pH i get 9.675
is this right?
You are exactly right. In fact, when I first responded to the problem I was not aware that the amount of NaOH added was enough that we no longer have a buffer but just sodium acetate salt, which of course, hydrolyzes. When I sat down to work the problem to explain what "you did wrong" I discovered what was going on. Your answer is right on the money.
To determine the pH of the solution, we need to consider the reaction between the acetic acid (CH3COOH) and sodium hydroxide (NaOH) in the buffer solution.
First, let's write the balanced equation for the reaction between acetic acid and sodium hydroxide:
CH3COOH + NaOH -> CH3COONa + H2O
From the balanced equation, we can see that one mole of acetic acid reacts with one mole of sodium hydroxide to produce one mole of sodium acetate and one mole of water.
Now, let's calculate the number of moles of acetic acid and sodium hydroxide present in the solution:
Number of moles of acetic acid = 2.00 mol/L * 0.0500 L = 0.1000 mol
Number of moles of NaOH = mass / molar mass = 4.000 g / 40.00 g/mol (molar mass of NaOH) = 0.1000 mol
Since the number of moles of acetic acid and sodium hydroxide is the same, they will react completely, and no excess of either will remain.
After the reaction, we are left with sodium acetate (CH3COONa) in the solution. Since sodium acetate is a strong electrolyte, it will dissociate completely in water, yielding sodium ions (Na+) and acetate ions (CH3COO-).
Now, let's calculate the concentration of sodium acetate in the solution:
Concentration of sodium acetate = 0.1000 mol / 0.0500 L = 2.00 mol/L
Since the final volume of the solution is still 50.00 mL, and the number of moles of sodium acetate is equal to the number of moles of acetic acid, the final concentration of sodium acetate is equal to the initial concentration.
Now, to determine the pH of the solution, we need to consider the dissociation of sodium acetate. Sodium acetate is the conjugate base of acetic acid, which means it acts as a weak base. When a weak base dissolves in water, it reacts with water to form hydroxide ions (OH-) and the corresponding acid (in this case, acetic acid).
However, since the concentration of sodium acetate is relatively high (2.00 M) compared to the concentration of water in the buffer solution, the effect of hydroxide ions on pH is negligible. Therefore, the pH of the solution will be determined mainly by the acetic acid, which is a weak acid.
To calculate the pH of the solution, we need to know the dissociation constant of acetic acid (Ka). The Ka value for acetic acid is 1.8 x 10^-5.
To simplify the calculations, we can use an approximation called the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
Where pKa is the negative logarithm of the dissociation constant (Ka), [A-] is the concentration of the conjugate base (sodium acetate), and [HA] is the concentration of the weak acid (acetic acid).
In our case, since the concentration of sodium acetate ([A-]) is equal to the initial concentration (2.00 M) and the concentration of acetic acid ([HA]) is the same, we can simplify the equation as follows:
pH = pKa + log (1/1) = pKa
Therefore, the pH of the solution is equal to the pKa of acetic acid, which is approximately 4.74.
Therefore, the pH of the solution prepared by adding 4.000 grams of NaOH to a 50.00 mL buffer that is 2.00 M in both acetic acid and sodium acetate is approximately 4.74.