Calculate the pH of the resultant solution of mixing together two solutions of 2 L of 1.0 M acetic acid, 0.5 L of 1.0 M sodium acetate and then adding 8 g of solid NaOH (MW = 40.0 g/mole).
Find starting # of moles of acid & base
(2L)(1M) = 2 moles HA
(.5L)(1M)= .5 moles A
Add 8 g NaOH
Convert grams to mole
8g (1mol/40g) = .2 moles NaOH
Find resulting # of moles
2 mol - .2 mol = 1.8 mol HA
.5 mol + .2 mol = .7 mol A
Use Henderson–Hasselbalch equation to find pH. pKa of acetic acid is 4.77
pH = pKa + log (A/HA)
pH = 4.77 + log (.7/1.8)
pH = 4.36
Your answer is 4.36!
To calculate the pH of the resultant solution, we need to determine the concentration of the acetic acid and acetate ions, and then consider the effect of adding solid NaOH. Here's how you can do it step by step:
Step 1: Determine the concentration of acetic acid (CH3COOH) and sodium acetate (CH3COONa) in the solution.
Given:
- 2 L of 1.0 M acetic acid solution
- 0.5 L of 1.0 M sodium acetate solution
The number of moles of acetic acid can be calculated using the formula:
moles of acetic acid = concentration (M) × volume (L)
In this case:
moles of acetic acid = 1.0 M × 2 L = 2 moles
The number of moles of sodium acetate is calculated similarly:
moles of sodium acetate = 1.0 M × 0.5 L = 0.5 moles
Step 2: Determine the excess OH- ions from the reaction of NaOH with acetic acid.
The balanced equation for the reaction of sodium hydroxide (NaOH) with acetic acid is:
CH3COOH + NaOH -> CH3COONa + H2O
From the equation, we know that 1 mole of NaOH reacts with 1 mole of CH3COOH to produce 1 mole of CH3COONa. Therefore, the number of moles of CH3COOH that will react with NaOH is 1 mole.
The number of moles of NaOH is calculated using the formula:
moles of NaOH = mass (g) / molar mass (g/mol)
In this case:
moles of NaOH = 8 g / 40.0 g/mol = 0.2 moles
Since NaOH reacts in a 1:1 ratio with acetic acid, only 0.2 moles of acetic acid will react. The remaining 1.8 moles (2 moles - 0.2 moles) of acetic acid are in excess.
Step 3: Determine the concentrations of acetic acid and acetate ions after the reaction.
The concentration of acetic acid after the reaction is calculated using the formula:
concentration (M) = moles / volume (L)
In this case:
concentration of acetic acid = 1.8 moles / 2 L = 0.9 M
Similarly, the concentration of acetate ions (CH3COO-) is calculated as:
concentration of acetate ions = moles of sodium acetate / total volume (L)
In this case:
concentration of acetate ions = 0.5 moles / (2 L + 0.5 L) = 0.5 moles / 2.5 L = 0.2 M
Step 4: Calculate the pH of the solution.
To calculate the pH, we need to consider the dissociation of acetic acid. Acetic acid is a weak acid, meaning it does not fully dissociate in water. It partially dissociates into hydrogen ions (H+) and acetate ions (CH3COO-).
The dissociation equation for acetic acid is:
CH3COOH ⇌ H+ + CH3COO-
The equilibrium constant for acetic acid (Ka) is approximately 1.8 x 10^(-5) at 25°C.
Using the equilibrium expression for Ka:
Ka = [H+][CH3COO-] / [CH3COOH]
We can assume that the [H+] concentration is equal to the concentration of acetic acid, as it is in much greater excess than the acetate ions from sodium acetate.
Substituting the concentrations we found in step 3:
1.8 x 10^(-5) = [H+][0.2 M] / [0.9 M]
Solving for [H+]:
[H+] ≈ 0.004 M
To convert [H+] to pH, we can use the formula:
pH = -log[H+]
pH = -log(0.004)
pH ≈ 2.40
Therefore, the pH of the resultant solution is approximately 2.40.