Question

The pH of an aqueous solution of 0.130 M sodium nitrite, NaNO2 (aq), is This solution i acidic basic neutral

Answer 1

Answer: solution is basic , because pH = 8.23 ,which is greater than 7.

explanation:

first writing down the equilibrium equation:

NO2- + H2O. <========> HNO2 + OH-

initial concentration of sodium nitrate = 0.130 M

We are constructing the ICE table.

	NO2-	HNO2	OH-
INITIAL	0.130	0	0
CHANGE	-x	+x	+x
EQUILIBRIUM	(0.130-x)	+x	+x

Ka for HNO2 = 4.5*10^-4

thus, kb = Kw/Ka = 1.0*10^-14 / 4.5 *10^-4

Kb = 0.222 *10^-10

Since , Kb = [HNO2][OH-]/ [NO2- ]

Kb = x*x/(0.130-x)

0.222*10^-10 = x2 / (0.130 - x )

x^2 +(0.222*10^-10) - (0.02886 *10^-10) = 0

Solving the quadratic equation, x = 1.69*10^-6 = [OH-]

pOH = -log [ OH-] = -log [ 1.69*10^-6] = 5.77

therefore, pH = 14-5.77 = 8.23 ,which is greater than 7.

thus , solution is basic

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