pH of an equivalence point

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Explanation & Answer

First find the molarity of the salt:

1 NaOH & 1 HNO2 --> 1 NaNO2 & H2O 

0.05 litres @ 0.25mol / litre HNO2 = 0.2 moles of HNO2 

0.2 moles of HNO2 needs 0.2 moles of NaOH 

0.2 moles of NaOH @ 1mol/litre = 0.2 litres of NaOH 

your NaNO2 : 0.2 mol / 0.25 litres = 0.8 Molar 
----------------------------------- 

the (NO2)- ion from the weak acid will do a hydrolysis with water, producing a basic solution : 

NO2- in water --> HNO2 & OH- 

the K hydrolysis = K water / K acid = 1e-14 / 5.1e-4 = 1.96e-11 

1.96e-11 = [HNO2] [OH-] / [NO2-] 

1.96e-11 = [x] [x] / [0.8] 

x2 = 1.568e-11

x = [OH-] = 3.98e-6 

pOH = 5.4

pH = 14 - pOH = 8.6

your answer: = 8.6


Anonymous
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