Guided Problem o.615 CH3cooA Calculate the pH at the equivalence point of 15 mL

Question

Guided Problem o.615 CH3cooA Calculate the pH at the equivalence point of 15 mL of 0.3 M acetic acid titrated with 0.2 M sodium hydroxide. For acetic acid, Ka = 1.8 x 105. Step 1: Determine the number of moles of base needed to titrate the acid to the equivalence point. moles 365)- o,) 0.00 93mo ??? ? 0-2 v000115 no!?.Na OH CO.RH)(00225) be ? Step 2: Determine the volume of base needed to titrate the acid to equivalence point oly CHscooua tou nesd to Use TCE,Table 0,00 tSmol 0.00 ton 6 Step 3: Determine the reaction and left over salts.

Explanation / Answer

1. Determine the number of moles of Base:

We first write out the reaction involved in the titration:

CH3COOH + NaOH -----> CH3COONa + H2O

Inorder to calculate the number of moles of base, we first have to calculate the moles of acid. Moles of acid can be calcualted using the molarity and volume of the acid.

Molarity of acetic acid = 0.3 M, Volume of acetic acid = 15ml = 0.015L

Moles of Acetic acid = Molarity* Volume = 0.3 moles/L * 0.015 L = 4.5 x 10-3 moles of acetic acid.

From the reaction, for every i mole of acetic acid, 1 mole of base is needed.

So moles of Base = 4.5 x 10-3 moles Acetic acid x (1mole of NaOH / 1 mole of acetic acid) = 4.5 x 10-3 moles NaOH.

Never use C1V1 = C2V2 for acid base titrations. Although you may luckily end up with correct moles, sometimes. It is strictly a formula used for dilutions.

2. Determine volume of the base:

Volume of the Base = Moles of Base/ Molarity of the base = (4.5 x 10-3 moles / 0.2 moles/L) = 0.0225 L

3. Determine the reaction and left over salts:

CH3COOH (aq) + NaOH (aq) -----> CH3COONa (aq) + H2O (l)

I   4.5 x 10-3 moles   4.5 x 10-3 moles   0 -

C - 4.5 x 10-3 moles   - 4.5 x 10-3 moles + 4.5 x 10-3 moles -

E 0    0 + 4.5 x 10-3 moles

4. Determine molarity of the new substances:

Molarity of CH3COONa = moles CH3COONa / Total Volume = 4.5 x 10-3 moles / (0.015+0.022) L = 0.122 M

*Total volume here is the volume of acid + volume of Base

5. Determine pH of the solution:

At the equivalence point we have a solution of sodium acetate. We havealready found out the concentration of this acetate in 4 as 0.122 M

CH3COO- + H2O ---------> CH3COOH + OH-

I0.122 M - 0 0

C-x - +x   +x

E 0.122-x - x x

So since sodium acetate is a conugate base of weak base,

kb = [OH-] [CH3COOH]/[CH3COONa]

Acetate is a weak base with ka = 1.8 x 10-5, we know that ka kb = 10-14=>, kb = 10-14 / (1.8 x 10-5) = 5.56 x 10-10

Substituting kb in the above equation:

5.56 x 10-10 = [OH-] [CH3COOH]/[CH3COONa]

5.56 x 10-10 = x2/(0.122 -x)

let us assume that x is small compared to 0.122 and ignore in the denominator

5.56 x 10-10 = x2/(0.122) => x2 = .122 x 5.56 x 10-10 = 6.78 x 10-11

x = sqrt (6.78 x 10-11) = 8.23 x 10-6

[OH-] = 8.23 x 10-6 => pOH = -log(8.23 x 10-6 ) = 5.08

pH = 14 - pOH = 14 - 5.08 = 8.92

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