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What mass of sodium nitrite must be added to 350 mL of water to give a...

What mass of sodium nitrite must be added to 350 mL of water to give a solution with a pH = 8.40? (Ka (HNO2) = 5.6 x 10-4)

Expert Solution

PH = 8.4

POH = 14-PH

= 14-8.4

= 5.6

-log[OH^-] = 5.6

[OH^-] = 10^-5.6 = 2.5*10^-6 M

Kb = Kw/Ka

= 1*10^-14/5.6*10^-4 = 1.78*10^-11

NaNO2(aq) --------------> Na^+ (aq) + NO2^- (aq)

NO2^- (aq) + H2O ----------------> HNO2(aq) + OH^- (aq)

I x 0 0

C -2.5*10^-6 2.5*10^-6 2.5*10^-6

E x-2.5*10^-6 2.5*10^-6 2.5*10^-6

Kb = [HNO2][OH^-]/[NO2^-]

1.78*10^-11 = 2.5*10^-6*2.5*10^-6/(x-2.5*10^-6)

1.78*10^-11 *(x-2.5*10^-6) = 2.5*10^-6*2.5*10^-6

x = 0.35

molarity = 0.35M

no of moles of NaNO2 = molarity *volume in L

= 0.35*0.35 = 0.1225 moles

mass of NaNO2 = no of moles * gram molar mass

= 0.1225*69 = 8.4525g >>>>>>>>answer

queen_honey_blossom answered 2 years ago

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