Question

What is the pH at the point of neutralization when 25.00 mL of 0.1000 molar acetic acid (Ka = 1.8 x 10-5) reacts with 25.00 mL of 0.1000 molar sodium hydroxide?

Answer 1

B) 8.72

Step-by-step explanation:

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Answer 2

The equation for the acid base reaction can be written as
CH3COOH + NaOH → CH3COONa + H2O
We calculate the number of moles of the acid which is equal to the base at equivalence point:
mol CH3COOH = 0.1000 M * 0.025L = 0.00250 mol
= mol H+ reacted = mol OH- used = mol CH3COO-
where the number of moles of salt CH3COONa formed is equal to the number of moles of acid and base.
Using the total volume of the solution, we now compute for the molarity of the salt solution:
[CH3COO-] = 0.00250 mol/(0.025L + 0.025L) = 0.0500 M
The equation for the hydrolysis reaction by the salt is given by
CH3COO- + H2O ⇔ CH3COOH + OH-
initial 0.0500 0 0
equilibrium 0.0500-x x x

Using Kw, we write an expression for Kb for the salt:
Kb (for CH3COO-) = Kw/Ka(for CH3COOH) = 1.0x10-14 / 1.8x10-5
Now solving for the concentration x of OH- formed:
Kb = x²/(0.0500-x)
approximating that x is negligible compared to 0.0500 simplifies the expression to
1.0*10^-14/1.8*10^-5 = x²/0.0500
x = 5.27*10^-6 = [OH−]
Finally, we calculate pH from pOH:
pOH = −log(5.27*10^-6) = 5.28
pH = 14−5.28 = 8.72