Print calculator Periodic Table uestion 6 of 12 General Chemistry 4th Edition Un

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Print calculator Periodic Table uestion 6 of 12 General Chemistry 4th Edition University Science Books presented by Saping Leaming Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.140 M HcIO(ag) with 0.140 MKOH(aq). The ionization constant for HCIO can be found here. abefore addition of any KOH 4.19 (b) after addition of 250 mL of KoH 7.537 click to edit (c) after addition of 30.0 mL of KOH (d) after addition of 50.0 mL of KOH There is a hirt availablel 0 mL of KOH divider bar again to hide the hint. Previous 8 Give up vew Solution Check Answer Next a Ext O about us I careers I partners privacy policy terms of use l contact us l help

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 KOH

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

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

[HA] = M – x

Substitute in Ka

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

Ka = x*x/(M-x)

3.5*10^-8 = x*x/(0.14 -x)

This is quadratic equation

x =6.99*10^-5

For pH

pH = -log(H+)

pH =-log(6.99*10^-5)

pH in a = 4.15

b) 25 ml KOH

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 = 50*0.14 = 7 mmol of acid

mmol of base = MV = 25*0.14 = 3.5 mmol of base

then, they neutralize and form conjugate base:

mmol of acid left = 7-3.5 = 3.5 mmol

mmol of conjguate left = 0 +3.5 = 3.5

Get pKa

pKa = -log(Ka)

pKa = -log(3.5*10^-8) = 7.45

Apply equation

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

pH = 7.45 + log (3.5/3.5) = 7.45

c) for 30 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)

mmol of acid = MV = 50*0.14 = 7 mmol of acid

mmol of base = MV = 30*0.14 = 4.2 mmol of base

then, they neutralize and form conjugate base:

mmol of acid left = 7-4.2 = 2.8 mmol

mmol of conjguate left = 0 +4.2     = 4.2   

Apply equation

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

pH = 7.45 + log (4.2/2.8) = 7.63

d) Addition of Same quantitie of Acid/Base

This will be Hydrolisis (equilibrium of acid-base) and the weak acid/base will form an equilibrium

We will need Kb

Ka*Kb = Kw

And Kw = 10^-14 always at 25°C for water so:

Kb = Kw/Ka = (10^-14)/(3.5*10^-8) = 2.857*10^-7

Now, proceed to calculate the equilibrium

H2O + A- <-> OH- + HA

Then K equilibrium is given as:

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

Assume [HA] = [OH-] = x

[A-] = M – x

Substitute in Kb

2.857*10^-7 = [x^2]/[M-x]

recalculate M:

mmol of conjugate = 20 mmol

Total V = V1+V2 = 50+50= 100

[M] = 7/100 = 0.07 M

2.857*10^-7 = [x^2]/[0.07-x]

x =1.41*10^-4

[OH-] =1.41*10^-4

Get pOH

pOH = -log(OH-)

pOH = -log 1.41*10^-4) = 3.85078

pH = 14-pOH = 14-3.85078= 10.14922

e) Addition of base

There will be finally an Excess of Base!

mol of acid < mol of base

Calculate pOH directly

[OH-] = M*V / Vt

mmol of acid = MV = 50*0.14= 7

mmol of base = MV = 60*0.14 = 8.4

therefre,

mmol of strng base left =8.4-7 = 1.4 mmol

Vtotal =50+60= 110 mL

[OH-] =1.4/110 = 0.127

pOH = -log(OH-)

pOH = -log(0.127) = 0.896

pH = 14-pOH = 14-0.896= 13.10

pH = 13.10

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