Answer to Question #140891 in General Chemistry for c

Question

A 0.10M solution of nitrous acid, HNO2, has an equilibrium hydrogen ion concentration of 6.7 X10-3M. what is the Ka of nitrous acid?

The Ka of benzoic acid, C6H5COOH is 6.3 X10-5 for C6H5COOH (aq) + H2O(l) ⇄ H+ (aq) + C6H5COO -(aq) What is the equilibrium [H+] in a 0.20M solution of benzoic acid?

If the hydrogen ion concentration is 10-8, what is the ph?

If the hydronium ion concentration is 2.5 X 10-4, what is the pH?

Answer 1

1) HNO₂ ↔ H⁺ +NO₂^-

[H⁺ ] = [NO₂^-] =  6.7×10^-3

Constant Ka = [H⁺] [NO₂^- ] / [HNO₂]

Ka = 6.7×10^-3 × 6.7×10^-3 / (0.10 – 6.7×10^-3) = 4.81×10^-4

2)Constant Ka = [H⁺] [C₆H₅COO^-] / [C₆H₅COOH]

[H⁺] = [C₆H₅COO^-] = x

Ka = 6.3×10^-5

[C6H5COOH] = 0.20 – x

If x is small, then you do not need to use the quadratic equation because you can assume (0.20 –x) = 0.20 The variable x is so small that it will not make much difference in  subtraction.

6.3×10^-5 = x² / 0.20

x = 3.5×10^-3

[H⁺] = 3.5×10^-3

3) pH = -log [H⁺ ]

pH = -log 10^-8 = 8

4) pH = 14 – pOH

pOH = -log [OH^-] = -log 2.5×10^-4 = 3.6

pH = 14 - 3.6 = 10.4

Answer: the Ka of nitrous acid = 4.81×10^-4

the equilibrium [H+] = 3.5×10^-3

pH = 8

pH = 10.4