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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