Question

What is the concentration of HOBr in a solution containing HOBr and NaOBr that has [NaOBr]=0.36M and pH=8.70 ? (Ka of HOBr=2.0 x 10^-9
Would I use the Henderssen-Hasselbach equation ?

Answer 1

yes, you would. you're just solving for something different this time.

Answer 2

I tried this and I ended up with a negative number so I'm having difficulty

Answer 3

so pH=pka +log [NaOBr]/[HOBr]
8.70= 8.60 +log [0.36]/x ?
Would this be the correct setup @aaronq?

Answer 4

I just took this chem exam and I believe this is the correct setup

Answer 5

Okay but when I solve I get -4.43 so I know I'm doing something wrong

Answer 6

8.70-8.60 = 0.10 , pKa would be 8.60
so 0.10 = log 0.36/x --> 0.10x=log 0.36 ?

Answer 7

no it's .10 = \[\log_{10} (\frac{ .36 }{ x })=.1\]

Answer 8

Oh okay, I'm getting confused with the log part, can you please explain. Thank you for helping me too

Answer 9

yessir this is back a bit so I forget it too but to rearrange the log you can set it up
\[10^{.1}=\frac{ .36 }{ x }\] and I'm sure you can solve it from there I get .286

Answer 10

Okay so just to check @PeachRings did you get 0.60 M as an answer ?

Answer 11

i got x to be .286 beccause 10^.1 is equal to 1.2589 and then take .36/1.2589

Answer 12

Oh because I'm given 4 answer choinces 0.60M , 0.36M, 0.14M and 0.98M and I don't see 0.285

Answer 13

well sheit lemme look

Answer 14

ahhh i had assumed your work was correct but the PKa of 2e-9 is .00103 so the answer is .36

Answer 15

Pka=-log(Ka)

Answer 16

Oh okay thank you so much