pH of an equivalence point

Chemistry
Tutor: None Selected Time limit: 1 Day

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

Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.
Click to visit
The Notebank
...
Apr 1st, 2015
...
Apr 1st, 2015
Feb 24th, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer