Helpp!! destroyer 2011 OC #178

[diene]

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I got this question correct but it's bugging me because I don't fully understand.

So what I did was find the pH, which is 9..pH 9 means that it's basic, so the pKa would have to be high. Based on that I eliminated choices and got the answer d, which is correct.

1.Can someone explain a better way of solving problems dealing with indicators, pH, and pKa?
2. if pH > pKa the solution will be deprotonated and pH < pKa, the solution will be protonated..is this correct?

Sorry for the bad pic quality..the hydroxide concentration is 1 x 10^-5 M if you can't read it properly from the pic.

Thanks!

 

[diene]

bumpp

 

[R1Guy]

I got this question correct but it's bugging me because I don't fully understand.

So what I did was find the pH, which is 9..pH 9 means that it's basic, so the pKa would have to be high. Based on that I eliminated choices and got the answer d, which is correct.

1.Can someone explain a better way of solving problems dealing with indicators, pH, and pKa?
2. if pH > pKa the solution will be deprotonated and pH < pKa, the solution will be protonated..is this correct?

Sorry for the bad pic quality..the hydroxide concentration is 1 x 10^-5 M if you can't read it properly from the pic.

Thanks!

On problems like this all you do is compare the ph of the solution to the pka of the indicator. If ph is less than pka, it will be protonated. If ph is greater than pka, it will be deprotonated.

So, the first thing we need to do is figure out the ph of the solution, and you must be able to do this mentally because the DAT loves to ask questions regarding acid/base chemistry. If the [OH-] is 1x10-5, the first thing you need to do is find pOH.
pOH = -log(1x10-5)
pOH= 5-log1 (this is how you do these mentally, so you must memorize log1 through log 9)
pOH= 5-0
pOH=5
pOH + PH=14
so ph=9...therefore it will be deprotonated in all the indicators.