Question

11- The pH of a sodium acetate-acetic acid buffer is 4.70. Calculate the ratio [CH3COO−] / [CH3COOH].






1- Calculate the pH of a 0.20 M NH3/0.20 M NH4Cl buffer after the addition of 25.0 mL of 0.10 M HCl to 65.0 mL of the buffer.


A- Enter your answer in the provided box.

A 21.0−mL solution of 0.110 M CH3COOH is titrated with a 0.240 M KOH solution. Calculate the pH after the following additions of the KOH solution:

(a) 0.00 mL
=_____


(b) 5.00 mL
=_______


C- Enter your answer in the provided box.

In a titration experiment, 21.9 mL of 0.813 M HCOOH neutralizes 24.4 mL of Ba(OH)2. What is the concentration of the Ba(OH)2 solution?
_____M

D- a) Calculate the pH of the 0.39 M NH3/ 0.73 M NH4Cl buffer system.

pH =


(b) What is the pH after the addition of 20.0 mL of 0.075 M NaOH to 80.0 mL of the buffer solution?

pH =



E- What mole ratio would you need to prepare a liter of "carbonate buffer" at a pH of 9.96?


______: 1.0 : 0

A) NaHCO3 : H2CO3 : Na2CO3
B) Na2CO3 : NaHCO3 : H2CO3

Choose the order of the compounds that is represented in the numeric ratio.



F- Enter your answer in the provided box.

The pH of a bicarbonate-carbonic acid buffer is 6.18. Calculate the ratio of the concentration of carbonic acid (H2CO3) to that of the bicarbonate ion (HCO3−).
(Ka1 of carbonic acid is 4.2 × 10−7.)

[H2CO3]
[HCO3−]
=

Answer 1

Answer 11

For a buffer solution, pH = pKa + log[sodium acetate]/[acetic acid]

pKa of acetic acid = 4.75

4.70 = 4.75 + log[sodium acetate]/[acetic acid]

10^-0.05 = [sodium acetate]/[acetic acid]

[sodium acetate]/[acetic acid]= 0.891

Answer A

When 0.00 mL of KOH is added, pH is due to acetic acid alone

pH = -log(K_a*C)^1/2

K_a is dissociation constant of acetic acid, C is the concentration

pH = -log(1.8 x 10^-5*0.11)^1/2 = 2.85

B. After addition if 5 mL of KOH

Moles of acetic acid present = 21 mL*0.11 M

= 2.31 millimoles

Moles of KOH added = 5 mL*0.24 M = 1.2 millimoles

As 1 mole of KOH reacts with 1 mole of acetic acid, remaining acetic acid = 2.31 - 1.2 = 1.11 millimoles

Molarity of remaining acetic acid = 1.11/26 = 0.043 M

pH = -log(K_a*0.043) = 3.05

molarity if Ba(OH)₂ can be calculated as follows

As 2 moles of HCOOH (acid) will react with 1 mole of Ba(OH)₂ (base)

Molarity of base = (Molarity of acid*Volume of acid)/(2*Volume of base) = (0.813*21.9 mL)/(2*24.4 mL) = 0.3648 M