An analytical chemist is titrating 63.1mL of a 0.3800M solution of acetic acid HCH3CO2 with a 0.8600M solution of KOH. The pKa of acetic acid is 4.70. Calculate the pH of the acid solution after the chemist has added 21.9mL of the KOH solution to it.Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added.Round your answer to 2 decimal places.
Question
An analytical chemist is titrating 63.1mL of a 0.3800M solution of acetic acid HCH3CO2 with a 0.8600M solution of KOH. The pKa of acetic acid is 4.70. Calculate the pH of the acid solution after the chemist has added 21.9mL of the KOH solution to it.Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added.Round your answer to 2 decimal places.
Solution
To solve this problem, we need to follow these steps:
-
Calculate the moles of acetic acid (HCH3CO2) and KOH initially present.
Moles of HCH3CO2 = Volume (L) x Molarity (M) = 0.0631 L x 0.3800 mol/L = 0.023978 mol
Moles of KOH = Volume (L) x Molarity (M) = 0.0219 L x 0.8600 mol/L = 0.018834 mol
-
Determine the limiting reactant. In this case, KOH is the limiting reactant because it has fewer moles than HCH3CO2. The reaction between HCH3CO2 and KOH will consume all of the KOH and some of the HCH3CO2, leaving some HCH3CO2 unreacted.
-
Calculate the moles of HCH3CO2 left after the reaction.
Moles of HCH3CO2 left = Moles of HCH3CO2 initially - Moles of KOH = 0.023978 mol - 0.018834 mol = 0.005144 mol
-
Calculate the concentration of HCH3CO2 left after the reaction.
Concentration of HCH3CO2 = Moles of HCH3CO2 / Total volume (L) = 0.005144 mol / (0.0631 L + 0.0219 L) = 0.0567 M
-
Use the Henderson-Hasselbalch equation to calculate the pH of the solution.
pH = pKa + log([A-]/[HA]) = 4.70 + log([0]/[0.0567]) = 4.70 + log(0) = 4.70 - ∞ = -∞
However, the pH of a solution cannot be negative or infinity. The log of zero is undefined, which means that the solution is fully acidic and the pH is very low. In reality, the pH would be determined by the autoionization of water and other factors. Therefore, the pH is approximately 0.
Similar Questions
An analytical chemist is titrating 102.5mL of a 0.2000M solution of formic acid H2CO2 with a 0.9800M solution of KOH. The pKa of formic acid is 3.74. Calculate the pH of the acid solution after the chemist has added 15.87mL of the KOH solution to it.Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added.Round your answer to 2 decimal places.
analytical chemist is titrating 55.5mL of a 1.100M solution of butanoic acid HC3H7CO2 with a 0.9700M solution of NaOH. The pKa of butanoic acid is 4.82. Calculate the pH of the acid solution after the chemist has added 69.8mL of the NaOH solution to it.Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of NaOH solution added.Round your answer to 2 decimal places.
A chemist titrates 50.0mL of a 0.4246M propionic acid HC2H5CO2 solution with 0.6989M NaOH solution at 25°C. Calculate the pH at equivalence. The pKa of propionic acid is 4.89.Round your answer to 2 decimal places.Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of NaOH solution added.
A 30.0 ml sample of 0.330 M propanoic acid (Ka = 1.3 x 10-5) is titrated with 0.600 M KOH. Calculate the pH after the addition of 20.5 mL of base.
Calculate the pH at 25°C of a 0.67M solution of potassium acetate KCH3CO2. Note that acetic acid HCH3CO2 is a weak acid with a pKa of 4.76.Round your answer to 1 decimal place.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.