Hi,
I am having trouble with two questions.
Find the pH of the following aqueous solutions prepared by adding:
a) 20mL 0.12M HCl to 10mL 0.16M CH3COOH + 20mL 0.20 KOH.
b) 10mL 5.0x10^-7 M HCl to 90mL H2O
For a), I thought it was a buffer solution so what I tried to do was find the initial moles and add or subtract the moles of HCl (as it reacts with the conjugate base to create more conjugate acid). But then I realised that CH3COOH and KOH were not conjugate pairs. So what do I do?
For b), I tried using M1V2 = M2V2 but the sheet said that "the trivial answer to this is not correct". So how do I do this ?
Thanks for your time.
For a, neutralize HCl with KOH (and there should be KOH left over) and use the excess KOH to neutralize CH3COOH. I think the mmoles left over from the first neutralization will be just the amount to neutralize the CH3COOH. So what you will have in the solutin is KCl (which doesn't affect the pH), H2O from both neutralizations (which is neutral), and CH3COOK (potassium acetate). The acetate ion )a base) is hydrolyzed like so
CH3COO^- + HOH ==> CH3COOH + OH^-
Kb = (K2/Ka) = (CH3COOH)(OH^-)/(CH3COOK).
Set up an ICE chart and substitute into the Kb equation and solve for OH^-, convert to pOH, then to pH. For CH3COOK, that will be the moles formed from the appropriate neutralization divided by the volume of ALL of the solutions.
For b, the trivial answer is not correct because the concn of the HCl given is so SMALL. Remember that the concn of H and OH in water is 1 x 10^-7 M. We usually neglect that when working with solutions BUT when the added soln is about the same, it is no longer negligible. So the total H^+ will be the sum of H^+ from HCl + H^+ from the H2O (which by the way will not be 1 x 10^-7 M).
Thanks,
But for the second one.
What will [H+] be if its not 1x10^-7 ?
For question a), you correctly recognized that CH3COOH and KOH are not a conjugate acid-base pair, so it is not a buffer solution. To find the pH of the solution, you can follow these steps:
1. Determine the moles of acid (HCl), base (KOH), and weak acid (CH3COOH) in the given volumes:
Moles of HCl = concentration (M) × volume (L) = 0.12 M × 0.02 L = 0.0024 moles
Moles of KOH = concentration (M) × volume (L) = 0.20 M × 0.02 L = 0.004 moles
Moles of CH3COOH = concentration (M) × volume (L) = 0.16 M × 0.01 L = 0.0016 moles
2. Determine the net ionic equation for the reaction between HCl and KOH:
HCl + KOH → KCl + H2O
3. Determine the moles of acid and base that react (if any) based on stoichiometry:
Based on the net ionic equation, 1 mole of HCl reacts with 1 mole of KOH. Since the moles of HCl and KOH are equal (0.0024 moles), they will completely react, leaving no excess acid or base.
4. Determine the remaining moles of CH3COOH after the reaction using the Henderson-Hasselbalch equation:
pH = pKa + log([conjugate base] / [weak acid])
For this equation, you need the pKa value of CH3COOH. Assume it is 4.75.
[conjugate base] is the moles of CH3COOH that reacted (0.0016 moles) + the moles of KOH that reacted (0.0024 moles) = 0.004 moles
[weak acid] is the initial moles of CH3COOH (0.0016 moles)
pH = 4.75 + log(0.004 / 0.0016) = 4.75 + log(2.5) ≈ 4.75 + 0.4 ≈ 5.15
Therefore, the pH of the solution is approximately 5.15.
Now for question b), the "trivial answer" mentioned on the sheet is likely referring to only using the dilution equation M1V1 = M2V2, which does not consider the ionization of HCl. To accurately find the pH of the solution, follow these steps:
1. Determine the concentration of HCl in the final solution:
Molarity (M) = moles of solute / volume of solvent (in liters)
Moles of HCl = concentration (M) × volume (L) = 5.0 × 10^(-7) M × 0.01 L = 5.0 × 10^(-9) moles
Volume of H2O = 90 mL = 0.09 L
Molarity of the final solution = 5.0 × 10^(-9) moles / 0.09 L ≈ 5.6 × 10^(-8) M
2. Use the formula for pH to calculate the pH of the solution:
pH = -log[H3O+]
[H3O+] is the concentration of H3O+ ions in the final solution. Since HCl is a strong acid and fully ionizes, the concentration of the H3O+ ions is equivalent to the concentration of the HCl in the final solution.
Therefore, pH = -log(5.6 × 10^(-8))
You can use a scientific calculator to find the final pH value by taking the logarithm of 5.6 × 10^(-8) and changing the sign.
Note: The value of pH will be negative; however, it should be noted that extreme pH values below 0 or above 14 are not commonly observed in everyday solutions.