ENCE 433 Dr. Alba Torrents

ENVIRONMENTAL ENGINEERING ANALYSIS Spring 2001

Homework Set # 6 In Class

ALKALINITY solutions

1.

a. 5x10^-4F NaOH

This is a strong base, [OH-] = [Na+] = 5 x 10^-4

pH = 10.7

Alk = 5 x 10^-4 eq/L = 25 mg/L as CaCO₃

b. 5x10^-5F Na₂CO₃

This is a weak base. The problem can be solved with a log C vs pH diagram and the following proton balance equation:

2[H2CO3] + [HCO3-] + [H^+] =[OH^-] which reduces to (as we add a base the fully protonated form can be neglected as is the [H^+]: [HCO3^-] =[OH^-] pH = 9.7 Alk = 10^-4 eq/L = 5 mg/L as CaCO3

c. 5x10-5F KHCO3

This is a weak base. The problem can be solved with a log C vs pH diagram and the following proton balance equation:

[H2CO3] + [H+] =[OH-]+ [CO3-2] which reduces to: [H2CO3] =[OH-] pH = 8 Alk = 5x10^-5 eq/L = 2.5 mg/L as CaCO3

d. 1x10^-3F HCl

This is a strong acid,

[Cl^-] = [H⁺] = 1 x 10^-3

pH = 3

Alk = -1 x 10^-3 eq/L = -50 mg/L as CaCO₃

2.

pH = 6.3

Alk = 1 x 10^-3 eq./L

Before equilibration with CO₂

After equilibration with CO₂ with P(CO₂) = 3.16x10^-4atm:

pH = ?

C_T = ?

Solve the alkalinity equation for open system:

Trial and error : pH = 8.3

C_T = 9.9x10^-4 M

C_T decreases and some CO₂ gets released to the atmosphere

3.

Alk + Acy = 2 C_T=10^-3M

C_T = 5x10^-4 M

Using the Deffeyes diagram: C_T = 0.5 mN

=======pH = 8.5

Alk = 0.5 mM

Using Equations:

4.

A. pH = 6.1

from diagram C_T = 1.3 mM

Alk = 0.5 mM

B. pH = 8.5

Not in diagram, solve equations

Alk = 2 mM

MIXTURE:

Diagram or by solving equations:

=====pH = 6.75

OPEN SYSTEM:

At pH = 6.75 ====== C_T = 10^-4.5 M

10^-4.5 M < 2x10^-3 M ===== Thus some CO₂ will evaporate