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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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s.h_math s.h_math
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6 months ago Edited: 6 months ago, s.h_math



What is the pH of a solution of 0.300 M HNO containing
0.190 M NaN02? (Ka of HN02 is 4.5 x 10-4)
1.02
Post Merge: 6 months ago

I got the answer for this one 3.15
Read 129 times
5 Replies
Replies
wrote...
#1
Educator
6 months ago
We have a similar question that goes like this:

Excerpt
What is the pH of a solution of 0.300 M HNO₂ containing 0.100 M NaNO₂? (Ka of HNO₂ is 4.5 × 10⁻⁴)

Solution:

Step-by-step explanation
Step 1:

The buffer consists of HNO2 a strong acid, and its conjugate base  NO2-, its conjugate base.

 

NaNO3(aq]→Na+(aq]+NO2-(aq]

 

[NaNO2]=[NO2-] = 0.1

[HNO⁠⁠⁠⁠⁠⁠⁠2] = 0.3

 

Step 2:

 

Ka = 4.5×10-4

 

pKa = - log (Ka)

= - log(4.5×10-4)

= 3.347

 

use:

 

pH = pKa + log {[conjugate base]/[acid]}

 

    = 3.347+ log {0.1/0.3}

    = 2.87

 

Answer:  2.87

 

Hope it was able to help you. If you have any doubt please tell me.

Thank you
s.h_math Author
wrote...
#2
6 months ago
Yes thank you, I had replied earlier to the post to tell you that I was able to answer it:)))


THANK YOU!
wrote...
#3
Educator
6 months ago
So is the answer 2.87 or 3.15 ?
s.h_math Author
wrote...
#4
6 months ago
3.15 to my question as it is slightly different
Answer accepted by topic starter
bio_manbio_man
wrote...
#5
Educator
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6 months ago
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