By how much the pH value changes when the amount of (NaOH) added varies from 24.

Question

By how much the pH value changes when the amount of (NaOH) added varies from 24.0mL to _______ From the above calculation, you will find that the solution's pH changes dramatically near the equivalence point. The phenolphthalein indicator changes its color (from colorless to pink) when pH > 8.2. not at pH = 7. If you perform a titration using phenolphthalein as the indicator, the titration end point is indicated by the color change of the indicator, so it happens at pH > 8.2 Do you think it is accurate to consider the titration end point as the same as the "equivalence point"? Why? Provide your answer to this question in your post laboratory. Now please consider what the pH of the solution is when 25.0mL of 0.01M NaOH is added to 25.0mL of 0, 01M acetic acid. Is it same as in the case of HCl solution? (Show your calculation procedure) What ph of the solution will be if exactly 25.0mL of 0.01M NaOH solution is added to 25.0mLof 0.01M of acetic acid solution? _________ Now let's find out what is the pH if half of the NaOH is added, in another words you have added 12.5mL of 0.01M NaOH solution to 25.0mL of 0.01M of acetic acid. There is a simple way to get the answer. You don't need to set up the mass conservation equilibrium. Think about what is the relationship between the concentration of acetate anion H_3C_2O_2:(aq) and acetic acid in the mixture when half of required NaOH is added. What pH of the solution will be if exactly 12.50mL of 0.01M NaOH solution is added to 25.0mL of 0.01M of acetic acid solution? ___________ After you completed the answers to the above questions, you now should have enough knowledge about how the solution pH changes during neutralization. You can proceed to the next part of the lab.

Explanation / Answer

Question 1.

Find pH of solution when V = 25 mL of NaOH, M = 0.1 M is added to V = 25 mL of 0.01 M of acetic acid

First, calculate mmol

mmol of acid = MV = 0.01*25 = 0.25 mmol of acetic acid

mmol of base = MV = 25*0.01 = 0.25 mmol of base

so, the base neutralizes the acid

there will be acetate formation

Acetat will hydrolyse to form:

A- + H2O <-> HA + OH-

Kb = [HA][OH-]/[A-]

For Kb,

Kb = Kw/KA = (10^-14)/(1.8*10^-5) = 5.55*10^-10

Recalculate [A-] = M1*V1/(V1+V2) = 25*0.01/(25+25) = 0.005 M

in equilibrium

[HA] = x = [OH-]

[A-] = M-x = 0.005 - x

solve for x

5.5*10^-10 = x*x/(0.005-x)

x = 1.665*10^-6

pOH = -log(1.665*10^-6) = 5.78

Question 2.

Find pH for:

mmol of acid = MV = 25*0.01 = 0.25 mmol of acid

mmol of base = 12.5*0.01 = 0.125

this is a buffer so

pH = pKa + log(A-/HA)

pKa = -log(1.8+*10^-5) = 4.75

After reaction:

mmol of A- formed = 0.125

mmol of HA left = 0.25 - 0.125 = 0.125

pH = 4.75 + log(0.125/0.125)

pH = 4.75

this is exactly half equivalence point

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