A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74 . Calculate the pH of this solution after 0.020

mol of NaOH is added.

To calculate the pH of the solution after adding 0.020 mol of NaOH, we need to determine how the solution components will react.

1. Firstly, let's determine the reaction that occurs when NaOH is added to the buffer solution:
NaOH + HC2H3O2 -> NaC2H3O2 + H2O

2. Calculate the concentrations of HC2H3O2 and NaC2H3O2 after the addition of NaOH:
Initially, we have 0.300 mol of both HC2H3O2 and NaC2H3O2 in 1.00 L of solution. This means the concentration of HC2H3O2 and NaC2H3O2 is 0.300 M.

Since NaOH reacts with HC2H3O2 to form NaC2H3O2, the moles of HC2H3O2 and NaC2H3O2 will be reduced by the same amount.

The moles of HC2H3O2 remaining after the reaction will be:
0.300 mol - 0.020 mol = 0.280 mol

The moles of NaC2H3O2 formed after the reaction will be:
0.020 mol

To calculate the new concentrations, we divide the number of moles by the volume of the solution:
HC2H3O2 concentration = 0.280 mol / 1.00 L = 0.280 M
NaC2H3O2 concentration = 0.020 mol / 1.00 L = 0.020 M

3. Calculate the pOH of the solution after the reaction:
Since NaOH is a strong base, the hydroxide ions (OH-) react with water to produce hydroxide ions (OH-) and increase the pH of the solution.

The pOH of the solution can be calculated using the concentration of OH- ions:
pOH = -log[OH-]

Since NaOH is a strong base, it completely dissociates in water, meaning the concentration of OH- ions is equal to the concentration of NaOH.

The concentration of OH- ions is:
0.020 mol / 1.00 L = 0.020 M

Therefore, the pOH of the solution is:
pOH = -log(0.020) = 1.70

4. Calculate the pH of the solution:
The pH and pOH of a solution are related by the equation: pH + pOH = 14

Therefore, the pH of the solution is:
pH = 14 - pOH = 14 - 1.70 = 12.30

Therefore, the pH of the solution after adding 0.020 mol of NaOH is 12.30.

To calculate the pH of a solution after adding NaOH to the buffer, we need to consider the reaction that occurs between the NaOH and the components of the buffer solution.

The reaction between NaOH and HC2H3O2 can be represented as follows:

NaOH + HC2H3O2 -> NaC2H3O2 + H2O

In this reaction, NaOH reacts with HC2H3O2 to form NaC2H3O2 (sodium acetate) and water.

Since the initial amounts of HC2H3O2 and NaC2H3O2 in the buffer solution are given as 0.300 mol each, the initial moles of HC2H3O2 and NaC2H3O2 remain the same even after NaOH is added.

After adding NaOH, the moles of NaC2H3O2 in the solution will increase, while the moles of HC2H3O2 remain unchanged. The total volume of the solution will also increase, but since we are dealing with a dilution of small amount (0.020 mol), the effect on pH will be negligible.

To calculate the new concentration of NaC2H3O2, we can use the concept of moles and volumes. The moles of NaC2H3O2 after adding NaOH can be calculated as follows:

Moles of NaC2H3O2 = Initial moles of NaC2H3O2 + Moles of NaOH

Moles of NaC2H3O2 = 0.300 mol + 0.020 mol

Moles of NaC2H3O2 = 0.320 mol

The total volume of the solution also changes. Initially, the volume was 1.00 L, and after adding NaOH, it will be slightly larger due to the slight increase in volume caused by the addition of NaOH.

Now, to calculate the new pH of the solution, we need to use the Henderson-Hasselbalch equation:

pH = pKa + log ([A-]/[HA])

In this equation, pKa is the logarithmic value of the acid dissociation constant of the weak acid (HC2H3O2), [A-] is the concentration of the acetate ion (NaC2H3O2), and [HA] is the concentration of the weak acid (HC2H3O2).

The pKa value for acetic acid (HC2H3O2) is given as 4.74 in the problem.

Let's substitute the values into the Henderson-Hasselbalch equation:

pH = 4.74 + log ([NaC2H3O2]/[HC2H3O2])

pH = 4.74 + log (0.320 mol/0.300 mol)

pH = 4.74 + log (1.07)

Using a calculator, we find that log(1.07) is approximately 0.03.

pH = 4.74 + 0.03

pH ≈ 4.77

Therefore, the pH of the solution after adding 0.020 mol of NaOH is approximately 4.77.

Not one, 4.68