  # MoeDent44

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## Everything posted by MoeDent44

1. My teacher said that my answer was correct. Thank you for helping out, you made me understand what the pKa and pKb does.
2. So I did this: nC2H5COOH = c * V = 0,10 mol/L * 0,025 L = 2,5 * 10^-3 mol nNaOH = c * V = 0,20 mol/L * 0,0125 L = 2,5 * 10 ^-3 mol When these two components react, we get: C2H5COOH + NaOH = C2H5COONa + H2O C2H5COONa = C2H5COO- + Na+ C2H5COO- + H20 = C2H5COOH + OH- Kb= 1,0*10^-14 (mol/l)^2 / Ka = 7,7*10^-10 mol/L [C2H5COOH] =c/Vtotal = 2,5*10^-3 mol / (0,0125 + 0,025) L = 0,0667 mol/L x^2=7,7*10^-10 * 0,0667 x^2 = 5,14 * 10^-11 x = 7,2 * 10 ^-6 mol/L OH- pOH = -log[OH-] = -log(7,2*10^-6) = 5,145 pH=14 - pOH = 14 - 5,145 = 8,86
3. Your help has necessary for me to understand more of the consept, but I'm still stuck, but thank you.
4. Thank you. The thing is that I've never encountered a task like this. I would know how to solve it if it were a strong base - strong acid titration, but this is a strong base - weak acid titration. I have the same amount of each solutions and I need to find the pH in the equivalence point. I also only have the concentration and the volume of C2H5COOH and the same of NaOH, which makes it difficult to use the Henderson Hasselbalch equation.
5. The Ka of C2H5COOH is 1,3*10^-5 mol/L. pKa is 3,8. How do I interpret this in the task? Is this my final answer?
6. Hey! So the task is: you have 25mL 0,10 mol/L CH3CH2COOH and 0,20 mol/L 12,5mL NaOH. What is the pH at the equivalence point. So I did as following: nCH3CH2COOH = c * V = 0,10 mol/L * 0,0250 L = 2,5 * 10^-3 mol nNaOH = c * V = 0,20 mol/l * 0,0125L = 2,5*10^-3 mol I don't know what to next, since this is a weak acid - strong base titration. If it were strong acid strong base and you had the same amount of mol, you'd get a pH of 7. I don't know what to do here.
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