If 20 mL of 0.10 M NaOH are added to 30 mL of 0.20 M HOAc, what is the pH of the resulting solution? (Ka = 1.8 ´ 10-5 for HOAc.)
Write the balanced equation, calculate the moles HOAc remaining and the moles NaOAc formed, then used the Henderson-Hasselbalch equation to solve for pH.
Thank you
To find the pH of the resulting solution, we need to consider the reaction between NaOH and HOAc, which will form sodium acetate and water.
First, we need to determine the amount of HOAc and NaOH that reacts. The balanced chemical equation for the reaction is:
HOAc + NaOH -> NaOAc + H2O
From the equation, we can see that the stoichiometry is 1 mole of HOAc reacts with 1 mole of NaOH. Since we know the molarity and volume of both substances, we can calculate the moles of HOAc and NaOH.
For HOAc:
moles of HOAc = (volume of HOAc in liters) * (molarity of HOAc)
moles of HOAc = (30 mL / 1000 mL/L) * (0.20 mol/L) = 0.006 moles
For NaOH:
moles of NaOH = (volume of NaOH in liters) * (molarity of NaOH)
moles of NaOH = (20 mL / 1000 mL/L) * (0.10 mol/L) = 0.002 moles
Since the stoichiometry is 1:1, we can see that 0.002 moles of NaOH will react with 0.002 moles of HOAc, leaving 0.004 moles of HOAc unreacted.
Now, we want to find the concentration (molarity) of the unreacted HOAc in the final solution. To do this, we need to determine the total volume of the resulting solution:
total volume = volume of HOAc + volume of NaOH
total volume = 30 mL + 20 mL = 50 mL = 0.050 L
To calculate the molarity of the unreacted HOAc, we divide the moles of HOAc by the total volume in liters:
molarity of unreacted HOAc = (moles of unreacted HOAc) / (total volume)
molarity of unreacted HOAc = 0.004 moles / 0.050 L = 0.080 M
Now, we need to calculate the concentration of the sodium acetate that formed. Since the reaction was 1:1, the concentration of sodium acetate is also 0.080 M.
Next, we need to calculate the concentration of H+ ions (acidity) in the solution. The sodium acetate that formed in the reaction can undergo partial hydrolysis, which results in the release of acetate ions (CH3COO-) and a small concentration of H+ ions. The acetate ions react with water to form hydroxide ions (OH-), according to the equation:
CH3COO- + H2O ⇌ CH3COOH + OH-
The Ka value for HOAc is given as 1.8 × 10^-5, which allows us to determine the concentration of H+ ions. We can set up an equilibrium expression:
Ka = [CH3COOH] * [OH-] / [CH3COO-]
Since the concentration of acetate ions is 0.080 M (the same as the unreacted HOAc), and we assume x is the small concentration of H+ ions:
Ka = x * x / 0.080
Solving the equation for x, we find that x = √(Ka * [CH3COO-]).
Substituting the given values:
x = √(1.8 × 10^-5 * 0.080) = 2.7 × 10^-3 M
Since the initial concentration of NaOH was 0.10 M, and all of it reacted, the concentration of OH- ions is also 0.10 M.
Now, we need to determine the concentration of H+ ions in the final solution:
[H+] = Kw / [OH-]
[H+] = (1.0 × 10^-14) / 0.10 = 1.0 × 10^-13 M
Finally, we can calculate the pH of the resulting solution using the formula:
pH = -log[H+]
pH = -log(1.0 × 10^-13)
pH ≈ 13.0
Therefore, the pH of the resulting solution is approximately 13.0.