Question

1) Write the hydrolysis reaction for the salt NaIO and determine if the solution is acidic, basic, or neutral.

2) A solution contains 0.050 M NaIO, what is ph of the solution? Ka(HIO)=2.3*10^-11

Answer 1

The hydrolysis reaction for the salt NaIO is:

NaIO + H2O → HIO + NaOH

In this reaction, the sodium ion (Na+) is a spectator ion and does not participate in the reaction. The iodate ion (IO3-) reacts with water to form the iodic acid (HIO) and hydroxide ion (OH-). Since the hydroxide ion is produced in the reaction, the solution is basic.

The dissociation of HIO in water can be written as follows:

HIO + H2O ⇌ H3O+ + IO3-

The equilibrium constant expression for this reaction is:

Ka = [H3O+][IO3-]/[HIO]

Since the concentration of NaIO is given, we can assume that the initial concentration of HIO is equal to the concentration of NaIO. Let x be the concentration of H3O+ and IO3- that is formed when HIO dissociates. Then, we have:

Ka = x^2 / (0.050 - x)

Assuming that x is much smaller than 0.050, we can simplify this expression to:

Ka = x^2 / 0.050

Solving for x, we get:

x = sqrt(Ka * 0.050) = 1.52 * 10^-6 M

The pH of the solution is given by:

pH = -log[H3O+] = -log(x) = 5.82

Therefore, the pH of the solution is 5.82