ENCE 433 Dr. Alba Torrents

EXAMINATION I "SOLUTION" Fall 1996

Tuesday, October 8

1.

(a) The molecular weight of CO3-2 is 60.

(b) The molecular weight of Ca3(PO4)2 is 310. Because each Ca3(PO4)2 forms six positive and six negative charges,

2.

We need to calculate the pH of the NaCN solution:

NaCN Na⁺ + CN^-

CN^- + H₂O === HCN + OH^-

Species H⁺, OH^-, Na⁺, CN^-, HCN

We have a base, so pH << 7.0

PBE: [OH^-] = [HCN] + [H⁺] ~ [HCN]

In the graph, intersection [OH^-] = [HCN] pH ~ 10.6

To have a pH = 6.0,

Initial number of moles of CN^- :

100,000 gal x 7.785 liters/gal x 10^-2 moles / liter = 3785 moles = C_T

After neutralization C_T = 3785 moles = 3783 moles HCN + 2 moles CN^-

We need to add 3783 moles of HCl = 3783 * 36.5 grams/mole = 138,079 grams

= 138,079 / 435.8 = 304.3 lb(15%)

3.

1. Strong base at a concentration > 10^-6 M

[OH^-] = 10^-3 M

pOH = 3 == pH = 11

1. Strong base at a concentration < 10^-6 M; We need to consider the dissociation of water.

ENE: [Na⁺] + [H⁺] = [OH^-] here it is not correct to assume [OH^- ] >>> [H⁺]

10^-8 + [H⁺] = K_w / [H⁺]

[H⁺] ² + 10^-8 [H⁺] - K_w = 0

[H⁺ ] = 9.51x10^-8 M

pH = 7.021 ~ 7

4.

At pH = 5 [HA] = 10^-3 M

[A^-] = 10^-8 M

[H⁺] = 10^-5 M

[OH^- ] = 10^-9 M

Addition of 10^-3 M A-

PBC: [HA] + [H⁺ ] = [OH^-] as we add a base, [OH^-] >>> [H⁺]

[HA] = ~ [OH^-] from graph pH ~ 10.5

Addition of 10^-3 HA

MB : 10^-3 = [HA] + [A^-]

ENE: [H⁺] = [A^-] + [OH^-]

K_a = [A^-] [H⁺] /[HA]

K_w = [H⁺] [OH^-] = 10^-14

As we add an acid, we can assume [OH^-] < [H⁺]

In ENE : [H⁺ ] = [A^-]

MBE: [A^-] = 10^-3 - [HA] or [HA] = 10^-3 - [A^-]

as is a weak acid and pH << 7 (we add acid) we can assume [HA] >>> [A^-] ~ 10^-3 M

[H⁺]² = K_a [HA] [H⁺]² = 10^-3 * 1,2* 10^-10 = 1,2*10^-13

[H⁺] = 3,46*10^-7 M pH = 6.46

(as pH < 6.5; there is a one log unit difference between [H⁺] and [OH^-] thus the error is less than 10% and the used assumption is OK. If a more accurate solution is needed, one would solve the system of equations without the assumption [OH^-] < [H⁺])