20 ml of 0.1 M Acetic acid and 25 ml of 0.1 M Sodium Acetate are mixed together. then 5ml of 0.1 M of NaOH is added. Find the Ph before the addition of NaOH and after the addition of NaOH??
Use the Henderson-Hasselbalch equation to find the initial pH.
Then complete the ICE chart I've started below for the second part and substitute the final mmoles into the HH equation. (which may help you on the first part, too).
begin = 20 mL x 0.1M = 20 mmoles CH3COOH.
begin = 25 mL x 0.1M = 25 mmoles CH3COONa
add 5mL x 0.1M NaOH = 5 mmoles OH-
.......CH3COOH + OH^- ==> CH3COO- + H2O
begin....20......0.........25.......0
add .............5...................
react....-5.....-5........+5........+5
final....
Iam not quite clear about how to find the Ph before the additon NaoH... could you please show me how to solve...
To find the pH before and after the addition of NaOH, we need to consider the acid-base reactions that occur in the solution.
First, let's determine the initial concentrations of acetic acid (CH3COOH) and sodium acetate (CH3COONa) in the mixture:
Initial volume of acetic acid = 20 ml
Initial molarity of acetic acid = 0.1 M
Initial moles of acetic acid = (20 ml) × (0.1 M) = 2 mmol
Initial volume of sodium acetate = 25 ml
Initial molarity of sodium acetate = 0.1 M
Initial moles of sodium acetate = (25 ml) × (0.1 M) = 2.5 mmol
Since the acetic acid is a weak acid, it will react with water to form hydronium ions (H3O+) and acetate ions (CH3COO-), according to the equation: CH3COOH + H2O ⇌ CH3COO- + H3O+
Given that the initial moles of acetic acid are 2 mmol, and the volume is 20 ml, the concentration of acetic acid is [CH3COOH] = (2 mmol) / (20 ml) = 0.1 M
The concentration of acetate ions is equal to the initial concentration of sodium acetate, which is 0.1 M, because sodium acetate completely dissociates into sodium ions (Na+) and acetate ions (CH3COO-).
Before the addition of NaOH, the concentration of hydronium ions in a solution of acetic acid can be calculated using the expression for the equilibrium constant (Ka) of the acid as follows:
Ka = [CH3COO-] × [H3O+] / [CH3COOH]
Since the concentration of acetate ions ([CH3COO-]) is equal to the initial concentration of sodium acetate, which is 0.1 M, and the concentration of acetic acid ([CH3COOH]) is also 0.1 M, the expression simplifies to:
Ka = [H3O+]
Solving for [H3O+], we find that the concentration of hydronium ions before the addition of NaOH is 0.1 M.
Therefore, the pH before the addition of NaOH can be calculated using the equation:
pH = -log[H3O+]
pH = -log(0.1) = 1
Now, let's consider the reaction that occurs after adding 5 ml of 0.1 M NaOH:
NaOH + CH3COOH → CH3COONa + H2O
Given the volume of NaOH added is 5 ml, the moles of NaOH added is (5 ml) × (0.1 M) = 0.5 mmol.
Since NaOH is a strong base, it completely dissociates into sodium ions (Na+) and hydroxide ions (OH-). Therefore, the concentration of OH- ions after the addition of NaOH is (0.5 mmol) / (30 ml) = 0.0167 M.
To calculate the concentration of hydronium ions ([H3O+]) after the reaction, we can use the fact that water reacts with hydroxide ions to form hydronium ions:
H2O + OH- → H3O+
Since the reaction is stoichiometric, the addition of 0.5 mmol of OH- ions will result in the formation of an equal amount of H3O+ ions. Therefore, the concentration of hydronium ions after the addition of NaOH is also 0.0167 M.
Converting the concentration of hydronium ions to pH:
pH = -log(0.0167) ≈ 1.78
Hence, the pH after the addition of NaOH is approximately 1.78.