Consider the titration of 100.0 mL of 0.260 M propanoic acid (Ka = 1.310−5) with 0.130 M KOH. Calculate the pH of the resulting solution after each of the following volumes of KOH has been added. (Assume that all solutions are at 25°C.)
(a) 0.0 mL
(b) 50.0 mL
(c) 100.0 mL
(d) 150.0 mL
(e) 200.0 mL
(f) 250.0 mL
To calculate the pH of the resulting solution after each volume of KOH has been added, we need to consider the reaction between the propanoic acid (CH3CH2COOH) and the KOH.
The balanced equation for the reaction is:
CH3CH2COOH + KOH → CH3CH2COOK + H2O
At the start of the titration (before any KOH is added), we have pure propanoic acid with a concentration of 0.260 M. This can be used to calculate the initial concentration of propanoic acid (CH3CH2COOH) and the initial concentration of propanoate ions (CH3CH2COO-) when the propanoic acid is in equilibrium with its conjugate base:
CH3CH2COOH ⇌ CH3CH2COO- + H+
Now, let's go through each step of the titration:
(a) 0.0 mL KOH added:
At this point, no KOH has been added, so the solution still contains only propanoic acid. To calculate the pH, we need to find the concentration of propanoic acid. The total volume of the solution is 100.0 mL, so the initial concentration of propanoic acid is 0.260 M. We can use the equation Ka = [CH3CH2COO-][H+] / [CH3CH2COOH] to find the concentration of H+ (which we can then convert to pH).
Since the Ka for propanoic acid is given as 1.310^-5, plug in the appropriate values into the equation:
1.310^-5 = [CH3CH2COO-][H+] / [CH3CH2COOH]
CH3CH2COO- = [H+] = x (assume x)
1.310^-5 = (x)(x) / 0.260
1.310^-5 = x^2 / 0.260
x^2 = 1.310^-5 * 0.260
x^2 = 3.406 * 10^-6
x = √(3.406 * 10^-6) = 0.001847 M (concentration of H+)
Now, calculate pH:
pH = -log[H+]
pH = -log(0.001847) = 2.73
So, at 0.0 mL of KOH added, the pH of the resulting solution is 2.73.
(b) 50.0 mL KOH added:
Now, 50.0 mL of 0.130 M KOH has been added. This will react with the propanoic acid in a 1:1 ratio. To find the concentrations of the reactants, we can use the following equation:
moles of CH3CH2COOH = initial moles - moles of KOH added
Initial moles of CH3CH2COOH = initial concentration * initial volume
= 0.260 M * 100.0 mL = 0.0260 moles
Moles of KOH added = concentration * volume
= 0.130 M * 50.0 mL = 0.0065 moles
Moles of CH3CH2COOH remaining = 0.0260 moles - 0.0065 moles = 0.0195 moles
Now, convert moles of CH3CH2COOH remaining to concentration:
Final volume = initial volume + volume of KOH added
= 100.0 mL + 50.0 mL = 150.0 mL
Final concentration = moles of CH3CH2COOH remaining / final volume
= 0.0195 moles / 150.0 mL = 0.013 M
To calculate the pH, we will repeat the steps from part (a) using the new concentration of CH3CH2COOH:
1.310^-5 = (x)(x) / 0.013
x^2 = 1.310^-5 * 0.013
x = √(1.310^-5 * 0.013) = 0.000114 M (concentration of H+)
pH = -log(0.000114) = 3.94
So, at 50.0 mL of KOH added, the pH of the resulting solution is 3.94.
You can repeat the same steps for the calculations of pH at each volume of KOH added (c, d, e, f) by adjusting the initial volume and concentrations based on the amounts of propanoic acid and KOH present at each step.