Answer to Question #170305 in General Chemistry for Mayuka

Question

A 0.10 M solution of formic acid (HCOOH) is prepared in various solvents:

a. pure water

b. 0.10 M NaHCOO solution,

c. 0.10 M Ca(HCOO)2 solution, and

d. 0.10 M NaCl.

Compute the pH of each solution. The Ka of formic acid is 1.78 x 10

Answer 1

A. In pure water

HCOOH :::::: HCOO- + H⁺

Let's find the equilibrium concentrations

[HCOOH]= 0.10-x

[H⁺] = x

[HCOO^-] = x

Ka= 1.78 x 10^-4

Now,

Ka= [HCOO-][H⁺]/[HCOOH]

Ka= x²/0.10-x

Since the acid is weak, 0.10-x is approximately 0.10

x²= 0.10 x Ka

x²= 0.10 x 1.78 x 10^-4

x= 4.22 x 10^-3M

[H⁺]= 4.22 x 10^-3M

pH= -log[H⁺]

pH= -log[4.22 x 10^-3]

pH= 2.37

B. NaHCO3 will ionize as follows

NaHCO3---> Na⁺ + HCO3^-

But HCO3^- will undergo hydrolysis

HCO3^- + H2O::::::: H2CO3 + OH^-

Lets find the equilibrium concentrations

[HCO3^-]= 0.10-x

[H2CO3]= x

[OH^-]= x

Ka= 4.8 x 10^-11

Ka= x²/0.10-x

Since HCO3^- is a weak acid, 0.10-x is approximately 0.10

x²= 0.10 x Ka= 0.10 x 4.8 x 10^-11

x= 2.19 x 10^-6

[OH^-] = 2.19 x 10^-6

[H⁺] = (4.22 x 10^-3) - (2.19 x 10^-6)

[H⁺]= 4.22 x 10^-3

pH=-log[H⁺]

pH= 2.37

This tells us that the pH remains fairly constant in NaHCO3

C. Ca(HCO3)2----> Ca²⁺ + 2HCO3^-

Then, HCO3^- will undergo hydrolysis. At equilibrium, the equilibrium concentrations are

[HCO3^-]= 0.20-x

[H2CO3]= x

[OH^-] = x

Ka= 4.8 x 10^-11

x²= 0.20 x Ka

x²= 0.20 x 4.8 x 10^-11

x²= 9.6 x 10^-12

x= 3.1 x 10^-6

[OH^-]= 3.1 x 10^-6

[H⁺] = (4.22x10^-3) - (3.1x10^-6)

[H⁺]= 4.189 x 10^-6

pH= -log[H⁺]

pH= -log[4.189 x 10^-6]

pH= 2.38

D. Since NaCl is neutral, the pH of formic acid in it will be the same as its pH in water.

pH= 2.37