ENCE 433 Dr. Alba Torrents

ENVIRONMENTAL ENGINEERING ANALYSIS Spring 2001

Homework Set #4 Out: March 1, 2001 Due: March 6, 2001

· 1.

· 2

NaOCl + H₂O ----------> Na⁺ + OCl^-

OCl^- + H₂O Ö HOCl + OH^-

Species: Na⁺, OCl^- , HOCl, OH^-, H⁺

MB : [OCl^- ] + [HOCl] = C_T = 10^-3

PBE: [HOCl] + [H⁺] = [OH^- ]

Assumption: as we add a base [OH^-] >>> [H⁺]

Thus in PBE : [HOCl] = [OH^- ]

In MB: [OCl^- ] = 10^-3 - [HOCl] = 10^-3 - [OH^- ]

At pH = 9.2, [HOCl] = [OH^- ]= 1.97x10^-5 M ;;;;;;;;;;

[OCl^-] = 1x10^-3 - 1.97x10^-5 ~ 1x10^-3 M

We need to add acid to reach pH = 7.5

At pH = 7.5, we can not follow the PBE as we do not know the type of acid added but:

At pK_a = pH

HOCl]= [OCl^-] � [HOCl]= [OCl^-]=CT/2 = 5.0 x 10^-4 M CT = [HOCl]= [OCl^-]

· 3.

Hydrochloric acid is an strong acid completely ionized in dilute aqueous solution. The self-ionization of water occurs at very slight extent in pure water, and in the presence of H₃O⁺ from the HCl, the self-ionization of water is repressed. This means that the reaction:

2 H₂O H₃O⁺ + OH^-

is driven to the left, because the reaction between HCl and H₂O produces and excess of H₃O⁺ . Thus

[H₃O⁺ ] = contribution from HCl + contribution from self ionization of water

= 0.400 + <<< 0.0000001 0.400

Using K_w we can calculate [OH^-]:

· 4

It is helpful first to classify the solutions as either acidic, basic, or neutral. There are three acidic solutions:

(a) 0.200 F NH₄NO₃ as NH₄⁺ ion is a weak acid

(d) 0.200 F HClO₄ perchloric acid is a strong acid

(f) 0.200 F HNO₂ nitrous acid is a weak acid

Three are two basic solutions:

(b) 0.200 F KNO₂ as NO₂^- ion is a weak base

(e) 0.200 F Na₂CO₃ as CO₃^-2 ion is a weak base

and the remaining solution, 0.200 F KNO₃, is a neutral solution. Neither K⁺ nor NO₃^- neither accept or donate protons to water. Another way of recognizing that is a neutral solution is to note that is produced by a reaction of an strong base (KOH) and a strong acid (HNO₃).

To decide on the relative acidity of the two weak acids, we must compare their K_a values:

K_a (HNO₂) = 4.5 x 10^-4

K_a (NH₄⁺) = 5.7 x 10^-10

indicating the nitrous acid is significant strongest than the ammonium ion.

To decide on the relative basicities of the two bases, we must calculate their K_b values from the K_a values of their conjugate acids:

Thus carbonate ion is a considerably stronger base than nitrite ion.

lowest pH, most acidic	(d) 0.200 F HClO4	strong acid
	(f) 0.200 F HNO2	Ka = 4.5 x 10^-4
	(a) 0.200 F NH4NO3	Ka = 5.7 x 10^-10
	© 0.200 F KNO3	neutral
	(b) 0.200 F KNO2	Kb = 2.2 x 10^-11
highest pH, most acidic	(e) 0.200 F Na2CO3	Kb = 2.1 x 10^-4

· 5

Species: H⁺, OH^-, HNO₂, NO₂^-

MBE: 0.1 = [NO₂^-] + [HNO₂]

ENE: [H⁺] = [NO₂^-] + [OH^-]

PBE: [H⁺] = [NO₂^-] + [OH^-]

As we add acid we can assume [H⁺] >>> [OH^-]

then from PBE: [H⁺] = [NO₂^-]

from MB : [HNO₂] = 0.1 - [H⁺]

In K_a [H⁺] ² = 10^-3.29 (0.1 - [H⁺])

[H⁺]² + 10^-3.29[H⁺] = 10^-4.29 = 0

pH = 2.16

Assumption [H⁺] >>> [OH^-] is OK as [OH^- ] = 10^-11.84

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