To achieve a pH of 4.1 in a solution of 450 ml of 2.5M acetic acid, approximately 43.88 grams of NaOH would need to be added. The percent equivalency after adding NaOH is 100%.
To calculate the grams of NaOH required to achieve a pH of 4.1 in a solution of acetic acid, we need to consider the acid-base reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH). This reaction produces water (H2O) and sodium acetate (CH3COONa).
1. Write the balanced chemical equation for the reaction:
CH3COOH + NaOH → H2O + CH3COONa
2. Determine the number of moles of acetic acid required to achieve a pH of 4.1:
Since acetic acid is a weak acid, we can assume that the pH is primarily determined by the concentration of acetic acid and its dissociation constant (Ka). The pH of 4.1 corresponds to a concentration of [H+] = 10^(-4.1) M.
Using the equation for the dissociation of acetic acid:
CH3COOH ↔ CH3COO- + H+
We can assume that the concentration of acetic acid is equal to the concentration of [H+], so the concentration of acetic acid is 10^(-4.1) M.
3. Calculate the moles of acetic acid:
Moles of acetic acid = volume (in liters) × concentration
Volume = 450 ml = 0.45 L
Moles of acetic acid = 0.45 L × 2.5 M = 1.125 moles
4. Since the stoichiometry of the balanced equation is 1:1, the moles of NaOH required will also be 1.125 moles.
5. Calculate the grams of NaOH:
Moles of NaOH = 1.125
Molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 39.00 g/mol
Grams of NaOH = Moles of NaOH × Molar mass of NaOH
= 1.125 moles × 39.00 g/mol
= 43.88 grams
Therefore, you would need to add approximately 43.88 grams of NaOH to achieve a pH of 4.1 in the solution of acetic acid.
To determine the percent equivalency, we need to compare the number of moles of NaOH used to the number of moles of acetic acid initially present.
6. Calculate the percent equivalency:
Percent equivalency = (moles of NaOH / moles of acetic acid) × 100
= (1.125 moles / 1.125 moles) × 100
= 100%