Got it

[atlanta213]

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In destroyer, # 17 of orgo, this is PH problem.

I don't see why they change to ka to kb to get the answer.

Since they ask PH, you can simply use ka and this is gonna give you same answer.

Some reason, destroyer always changed to ka or kb according to OH- formed or H+.

I don't see why we need to do this.

If i am wrong, somebody please explain this.

 

[Bruinlove]

Some reason, destroyer always changed to ka or kb according to OH- formed or H+.

Ka is acid dissociation constant while Kb is base dissociation constant.
Use Kb when OH- forms, Ka when H+ or H3O+ forms.
When you use Kb, you will end up with [OH-] and pOH. Calculate pH by 14-pOH.

Seriously though, you need to take basic G. Chem course before studying for DAT.

 

[atlanta213]

Ka is acid dissociation constant while Kb is base dissociation constant.
Use Kb when OH- forms, Ka when H+ or H3O+ forms.
When you use Kb, you will end up with [OH-] and pOH. Calculate pH by 14-pOH.

Seriously though, you need to take basic G. Chem course before studying for DAT.

Man I know kb corresponds to OH- precipitate and Ka to H+ precipitate.

The question that I asked was I don't see point of switching ka to kb for that specific problem since I could get the answer by the other way.

 

[Bruinlove]

Man I know kb corresponds to OH- precipitate and Ka to H+ precipitate.

The question that I asked was I don't see point of switching ka to kb for that specific problem since I could get the answer by the other way.

I don't know which problem you are talking about, but are you saying that the book uses Kb even though H+ forms?

 

[atlanta213]

kb for OH-
ka for H+

🙂

 

[futuredent22]

The reaction is

-CN + H2O -> HCN + OH

If you want to find the pH of a .04 M NaCN solution, you first need to solve for [OH] and then find pOH. To do that, you need to use the Kb of -CN.

 

[Bruinlove]

-CN + H2O -> HCN + OH

Hey OP, how do you use Ka to find pH of that reaction? If I understood you right, you said you could use Ka to get the same answer.

 

[atlanta213]

ok i will explain,

According to destroyer explanation, they changed to kb since OH- is formed when react with water. After that, they find POH from that calculation and 14-pOH to get an answer.

This is my way of solving this problem.
Since the question ask PH. Basically you times (0.040M) CN- to (5.0 x 10^-10) Ka to get H+ ([HCN][H+])/(CN-)

Then log of what you got. -log[2 x 10^-11]

Since it is close to 1 X 10^-11, the PH would be around 11.

Basically same thing, but i just wanna mentioned that this problem can be done without changing Ka to Kb.

 

[atlanta213]

I knew Ka, kb and what those two value indicates. 😡

I just wanted to introduce another way to tackle this one.

You discouraged me.