What is the pH of a 0.50 M solution of nitrous acid with a Ka of...

What is the pH of a 0.50 M solution of nitrous acid with a Ka of 4.0 x 10^-4?

Solutions

Expert Solution

Nitrous acid = HNO₂

HNO₂ H⁺ + NO₂^-

IC 0.5 0 0

EC 0.5-x x x

ka = 4.0 x 10^-4

Now,

ka = [H+][OH-]/[HNO2]

4.0 x 10^-4 = x² / (0.50-x)

2.0 x 10^-4 – (4.0 x 10^-4) x = x²

x² + (4.0 x 10^-4) x - (2.0 x 10^-4) x = 0

Solving the quadratic equation, we get

x= 0.0139 and x= -0.0143

Neglecting the negative value of x, we get

x = 0.0139

x = [H+] = [NO2-] 0.0139 M

pH = -log[H+] = -log(0.0139 ) = 1.86

queen_honey_blossom answered 1 year ago

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