Question

What is the pH of 65 mL of 0.0685 M HNO2?

Answer 1

Consider dissociation of HNO ₂ in water as, HNO _2(aq) + H₂O _(l) H₃O⁺_(aq) +NO ₂^-_(aq)

The expression for dissociation constant is given as, Ka = [H₃O⁺] [NO ₂^- ]/ [ HNO ₂ ] = 7.1 $\times$ 10 ^-04

To find out equilibrium concentration, let's use ICE table.

Concentration (M)	HNO 2	H3O+	NO 2-
Initial	0.0685
Change	-X	+X	+X
Equilibrium	0.0685-X	X	X

$\therefore$ Ka = (X)(X) / 0.0685 - X = 7.1 $\times$ 10 ^-04

X ² / 0.0685 - X = 7.1 $\times$ 10 ^-04  

X ² = 7.1 $\times$ 10 ^-04 (0.0685 - X)

X ² = 4.86 $\times$ 10 ^-05 - 7.1 $\times$ 10 ^-04 X

X ² + 7.1 $\times$ 10 ^-04 X - 4.86 $\times$ 10 ^-05 =0

Compare above equation with a X ² + b X+ c = 0, we get a = 1, b = 7.1 $\times$ 10 ^-04 and c= - 4.86 $\times$ 10 ^-05

Now, solving above equation for X, we get

X = -b +/- $\sqrt{}$( b ² - 4 a c ) / 2a

X = -  7.1 $\times$ 10 ^-04 +/- $\sqrt{}$(7.1 $\times$ 10 ^-04) ² - 4 (1)(- 4.86 $\times$ 10 ^-05) / 2 (1)

X = -  7.1 $\times$ 10 ^-04 +/- $\sqrt{}$5.04 $\times$ 10 ^-07 + 1.944  $\times$ 10 ^-04 / 2

X = -  7.1 $\times$ 10 ^-04 +/- $\sqrt{}$1.949  $\times$ 10 ^-04 / 2

X = -  7.1 $\times$ 10 ^-04 +/- 0.01396 / 2

Therefore, X =-  7.1 $\times$ 10 ^-04 + 0.01396 / 2   = 6.625 $\times$ 10 ^-03

Or X = -  7.1 $\times$ 10 ^-04 - 0.01396 / 2 = - 7.34 $\times$ 10 ^-03

Acceptable value of X is 6.625 $\times$ 10 ^-03 .

X = 6.625 $\times$ 10 ^-03 M = [H₃O⁺] = [NO ₂^- ]

We have, pH = -log [H₃O⁺] = - log (6.625 $\times$ 10 ^-03) = 2.18

ANSWER : pH of 0.0685 M HNO ₂ solution = 2.18