pH of an equivalence point

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Apr 1st, 2015

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

Apr 1st, 2015

It says its wrong so can u please recheck it.

Apr 2nd, 2015

Hey fufu, sorry for late response, can you try to type 8.35 as the answer? Let me know, thanks. 

Apr 2nd, 2015

thanks that helped

Apr 2nd, 2015

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