You add 12.5 mL of 4.15 acetic acid to 25.0 mL of 1.00 M NaOH. Calculate the hydronium ion concentration and pH of the resulting solution.
To calculate the hydronium ion concentration and pH of the resulting solution after the reaction between acetic acid and NaOH, we need to consider the balanced chemical equation and the stoichiometry of the reaction.
The balanced chemical equation for the reaction can be written as follows:
CH3COOH + NaOH -> CH3COONa + H2O
From the equation, we see that 1 mole of acetic acid (CH3COOH) reacts with 1 mole of NaOH to form 1 mole of water (H2O). Therefore, the mole ratio of CH3COOH to H2O is 1:1.
Given that you added 12.5 mL of 4.15 M acetic acid, we can calculate the number of moles of acetic acid using the formula:
moles of acetic acid = (volume of acetic acid in liters) × (concentration of acetic acid in M)
First, convert the volume of acetic acid from milliliters (mL) to liters (L):
volume of acetic acid = 12.5 mL ÷ 1000 = 0.0125 L
Now, we can calculate the number of moles of acetic acid:
moles of acetic acid = (0.0125 L) × (4.15 M) = 0.051875 mol
Since the mole ratio of acetic acid to water is 1:1, the number of moles of water formed is also 0.051875 mol.
Therefore, the hydronium ion concentration ([H3O+]) is equal to the number of moles of hydronium ions divided by the total volume of the solution. The total volume of the solution is the sum of the volumes of acetic acid and NaOH used.
Total volume of solution = volume of acetic acid + volume of NaOH = 12.5 mL + 25.0 mL = 37.5 mL = 0.0375 L
Since the mole ratio of NaOH to water is also 1:1, the number of moles of NaOH used is 0.051875 mol.
Therefore, the hydronium ion concentration is:
[H3O+] = (moles of hydronium ions) ÷ (total volume of solution) = (0.051875 mol) ÷ (0.0375 L) = 1.381 M
To calculate the pH of the solution, we can use the formula:
pH = -log[H3O+]
Substituting the value of [H3O+], we have:
pH = -log(1.381) ≈ 0.86
Therefore, the hydronium ion concentration of the resulting solution is approximately 1.381 M, and the pH is approximately 0.86.