Question

Consider the titration of a 25.0 mL sample of 0.110 molL−1 CH3COOH (Ka=1.8×10−5) with 0.130 molL−1 NaOH. Determine each quantity:

Part A

Part complete

the initial pH

Express your answer using two decimal places.

pH =

2.85

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Correct

Part B

Part complete

the volume of added base required to reach the equivalence point

V =

21.2

mL

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Correct

Part C

Part complete

the pH at 4.0 mL of added base

Express your answer using two decimal places.

pH =

4.11

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Correct

Part D

Part complete

the pH at one-half of the equivalence point

Express your answer using two decimal places.

pH =

4.74

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Correct

Part E

Part complete

the pH at the equivalence point

Express your answer using two decimal places.

pH =

8.76

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Correct

Part F

the pH after adding 5.0 mL of base beyond the equivalence point

Express your answer using two decimal places.

CAN SOMEONE ANSWER QUESTION F ONLY PLEASE!!!!!!!!

Answer 1

F)

Given:

M(CH3COOH) = 0.11 M

V(CH3COOH) = 25 mL

M(NaOH) = 0.13 M

V(NaOH) = 5 mL more than equivalence point

= 5 mL + 21.2 mL

= 26.2 mL

mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)

mol(CH3COOH) = 0.11 M * 25 mL = 2.75 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.13 M * 26.2 mL = 3.406 mmol

We have:

mol(CH3COOH) = 2.75 mmol

mol(NaOH) = 3.406 mmol

2.75 mmol of both will react

excess NaOH remaining = 0.656 mmol

Volume of Solution = 25 + 26.2 = 51.2 mL

[OH-] = 0.656 mmol/51.2 mL = 0.0128 M

use:

pOH = -log [OH-]

= -log (1.281*10^-2)

= 1.8924

use:

PH = 14 - pOH

= 14 - 1.8924

= 12.1076

Answer: 12.11