Question

A. Find "[OH^-]= ____ M" for 5.0x10-2 M NaBrO.

B. Find the pH for part A.

C. Find [OH^-] = ___ M" for a mixture that is 0.12 M in NaNO₂ and 0.15 M in Ca(NO₂)₂.

D. Find the pH for part C.

Answer 1

a) The reaction that occurs is:

BrO- + H2O = HBrO + OH-

The Kb of that reaction is 5x10 ^ -6

of the Kb expression:

Kb = [HBrO] * [OH-] / [BrO-]

we replace:

5x10 ^ -6 = X ^ 2 / 0.05 - X

we assume that - X is negligible and we have:

X = [OH-] = √5x10 ^ -6 x 0.05 = 3.24x10 ^ -6 M

b) Calculate pOH and pH:

pOH = - log [OH-] = - log (3.24x10 ^ -6) = 3.30

pH = 14 - 3.3 = 10.70

c) The total concentration of NO2- is calculated:

[NO2-] total = 0.12 M + 2 * 0.15 M = 0.42 M

The same procedure is applied above, for the reaction:

NO2- + H2O = HNO2 + OH-

Where Kb = 2.5x10 ^ -11

The concentration of OH-:

[OH-] = √2.5x10 ^ -11 * 0.42 = 3.24x10 ^ -6

d) Calculate pOH and pH:

pOH = - log (3.24x10 ^ -6) = 5.49

pH = 14 - 5.49 = 8.51

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