What is the pH when 25 mL of .20 M CH3COOH has been titrated with 35 mL of .10 M NaOH?
The pH is determined by the hydrolysis of the salt, CH3COONa.
CH3COO^- + HOH ==> CH3COOH + OH^-
Kb = (Kw/Ka) = (CH3COOH)(H3O^+)/(CH3COO^-)
Set up an ICE chart and substitute into the Kb equation. Solve for H3O^+ and convert to pH.
To find the pH when 25 mL of 0.20 M CH3COOH has been titrated with 35 mL of 0.10 M NaOH, we can use the Henderson-Hasselbalch equation.
The Henderson-Hasselbalch equation is given as:
pH = pKa + log ([A-]/[HA])
where pH is the unknown pH value, pKa is the dissociation constant of the weak acid (CH3COOH), [A-] is the concentration of the conjugate base of the weak acid (CH3COO-), and [HA] is the concentration of the weak acid (CH3COOH).
To use this equation, we need to determine the concentrations of the weak acid and its conjugate base.
Given the initial volume and concentration of CH3COOH (25 mL, 0.20 M) and the volume and concentration of NaOH used in the titration (35 mL, 0.10 M), we can calculate the moles of CH3COOH and NaOH.
Moles of CH3COOH = volume (in liters) × concentration of CH3COOH
= 0.025 L × 0.20 M
= 0.005 moles
Moles of NaOH = volume (in liters) × concentration of NaOH
= 0.035 L × 0.10 M
= 0.0035 moles
Next, we need to determine the limiting reagent in the reaction between CH3COOH and NaOH. Since the stoichiometric ratio between CH3COOH and NaOH is 1:1 (1 mole of CH3COOH reacts with 1 mole of NaOH), we can see that NaOH is the limiting reagent because we have fewer moles of NaOH than CH3COOH.
Since NaOH is the limiting reagent, it will react completely with CH3COOH to form the conjugate base, CH3COO-.
After the reaction is complete, the concentration of CH3COOH will be reduced by the moles of NaOH used, and the concentration of CH3COO- will be equal to the moles of NaOH used since CH3COOH reacts in a 1:1 mole ratio with NaOH.
Now, we can calculate the new concentrations of CH3COOH and CH3COO- after the reaction.
Concentration of CH3COOH = (moles of CH3COOH - moles of NaOH) / volume (in liters)
= (0.005 moles - 0.0035 moles) / 0.025 L
= 0.06 M
Concentration of CH3COO- = moles of NaOH / volume (in liters)
= 0.0035 moles / 0.035 L
= 0.10 M
Therefore, the concentration of CH3COOH is 0.06 M, and the concentration of CH3COO- is 0.10 M.
Finally, we can substitute these values into the Henderson-Hasselbalch equation to find the pH:
pH = pKa + log ([A-]/[HA])
= pKa + log (0.10/0.06)
Since CH3COOH is a weak acid, it dissociates partially into its conjugate base CH3COO-. Hence, we can assume that the concentration of CH3COOH is approximately equal to the initial concentration of CH3COOH (0.20 M) because the reaction proceeds to a small extent. Therefore, we can use the pKa value of acetic acid (the dissociation constant of CH3COOH), which is 4.76.
pH = 4.76 + log (0.10/0.06)
Calculating this expression, we find:
pH ≈ 4.76 + log (1.67)
Using logarithm properties, we can evaluate the expression inside the logarithm:
log (1.67) ≈ 0.22
Finally, substituting this value into the pH equation:
pH ≈ 4.76 + 0.22
pH ≈ 4.98
Therefore, the pH when 25 mL of 0.20 M CH3COOH has been titrated with 35 mL of 0.10 M NaOH is approximately 4.98.