ph
amt of (nh4)2so4 to be added to 500 ml 0.1M nh4oh to prepare a buffer of ph 9.56. pka of nh4+ is 9.26
where am i going wrong
18 Replies
@Dexter
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+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.can you try this from first principles instead of using the formula ek baar?
Example of first principles: https://youtu.be/q1Kh5Ls9WRM?list=PL7AAT-ai0VD4uk4r5AOYzGfRfahip-SKA&t=2340
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i get pretty much the same ig
idk my error
Can you share the working? And also the original q?
0.01 /0.05 ka log is -0.7
Putting what you said in the equation does not give you 9.56
If you are doing something quickly, gotta cross check like this too :)
I did some approximations
-1.9= log 2x/0.05 was created iirc
ah cool should work then!
ans is given 0.025 tho
dekh bhai ratio aata hai 0.3
which is log 2
if you are going to approximate when it is not needed risky hai :D
isiliye bola do on pen and paper properly
or if not, use calci
ignore the math i did xD please do this bit that anti log 0.3 is 2
(Edit ulta likha tha upar. My b)
usse ho jayega mujhe lagta hai
kiska ratio
can u pls do it on pen and paper and verify which ans is correct
XD.
This is what I am requesting you to do
I have sent my pen paper soln upar
But it is not right. The substitution.
Padh le upar bhai
Log ka calculator se kar :D one needs to be solid with math tools to make ionic calc work.
1. See your hand written last step.
2. Use calc to do this.
And please close this out post that :)
Yep. That’s the answer. 0.0125
ig the best approach is to take -0.3 as log (1/2)
gives exactly 0.0125
4 decimal places
anyways thanx
0.025 is confirmed wrong then
+solved @iTeachChem
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