Hi this is a question with many parts, thank you so much to anyone who can help

Question

Hi this is a question with many parts, thank you so much to anyone who can help out!

Part A: Titration of a weak acid

Calculate the pH during the titration of 10.00 mL of 0.400 M hypochlorous acid with 0.500 M NaOH. First what is the initial pH (before any NaOH is added)?

The Ka for HOCl is 3.0 x 10-8 M.

Part B: How many mL of NaOH are added to reach the equivalence point?

Part C: What is the pH after 3.50 mL of NaOH are added?

Part D: What is the pH after 4.70 mL of NaOH are added?

Part E: What is the pH at the equivalence point? (Notice that the pH of a weak acid at the equivalence point is basic.)

Part F: What is the pH after 9.00 mL of NaOH are added?

Part G: What is the pH when half the acid has been neutralized?

Part H: What is the pH after 20.30 mL of NaOH are added?

Explanation / Answer

First of all, volumes will change in every case (Volume of base + Volume of acid), therefore the molarity will change as well

M = moles / volume

pH = -log[H+]

a) HA -> H+ + A-

Ka = [H+][A-]/[HA]

a) no NaOH

Ka = [H+][A-]/[HA]

Assume [H+] = [A-] = x

[HA] = M – x

Substitute in Ka

Ka = [H+][A-]/[HA]

Ka = x*x/(M-x)

3*10^-8 = x*x/(0.4-x)

This is quadratic equation

x =1.095*10^-4

For pH

pH = -log(H+)

pH =-log(1.095*10^-4)

pH in a = 3.96

b)

mmol of acid = mmol of base

Macid*Vacid = Mbase * Vbase

Vbase = Macid*Vacid /Mbase

Vbase = 0.4*10/0.5 = 8 mL of base required

c) 3.5 ml KOH

This is in a Buffer Region, a Buffer is formed (weak base and its conjugate)

Use Henderson-Hassebach equations!

NOTE that in this case [A-] < [HA]; Expect pH lower than pKa

pH = pKa + log (A-/HA)

initially

mmol of acid = MV = 10*0.4 = 4mmol of acid

mmol of base = MV = 3.5*0.5= 1.75 mmol of base

then, they neutralize and form conjugate base:

mmol of acid left = 4-1.75 = 2.25 mmol

mmol of conjguate left = 0 + 1.75 = 1.75

Get pKa

pKa = -log(Ka)

pKa = -log(3*10^-8) = 7.52

Apply equation

pH = pKa + log ([A-]/[HA]) =

pH = 7.52+ log (1.75/2.25) = 7.41

d) for 4.70 ml

This is in a Buffer Region, a Buffer is formed (weak base and its conjugate)

Use Henderson-Hassebach equations!

pH = pKa + log (A-/HA)

initially

mmol of acid = MV =4 mmol of acid

mmol of base = MV = 4.7*0.5 = 2.35 mmol of base

then, they neutralize and form conjugate base:

mmol of acid left = 4-2.35 = 1.65 mmol

mmol of conjguate left = 0 + 10= 2.35

Apply equation

pH = pKa + log ([A-]/[HA]) =

pH = 7.52 + log (2.35 /1.65 ) = 7.67

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