2016 DAT Destroyer Questions

[blue222]

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I have a question about the explanations of questions 255 and 256 of the chemistry section of the 2016 DAT Destroyer.

For 255:

Is the pka value determined using the Henderson-Hasselbach eqn as such: 9 = pka + log [ (1x10^-5)/ (1x10^0)]? and if so, is the HA 1x10^0 because the question does not indicate any acid added to the solution? I initially assumed the given [OH] was at the equivalence point but would this be incorrect because there is no equal amount of acid?


For 256:

I guessed the pH would be around 5 and therefore got the right answer. However the solution explanation does not explain why the pH would be 5 or 6. Is this a fact that should be memorized or was this derived somehow?

 

[DAT Destroyer]

I have a question about the explanations of questions 255 and 256 of the chemistry section of the 2016 DAT Destroyer.

For 255:

Is the pka value determined using the Henderson-Hasselbach eqn as such: 9 = pka + log [ (1x10^-5)/ (1x10^0)]? and if so, is the HA 1x10^0 because the question does not indicate any acid added to the solution? I initially assumed the given [OH] was at the equivalence point but would this be incorrect because there is no equal amount of acid?


For 256:

I guessed the pH would be around 5 and therefore got the right answer. However the solution explanation does not explain why the pH would be 5 or 6. Is this a fact that should be memorized or was this derived somehow?

For 255, The pH is determined using the equation pH +pOH = 14. Since you have the OH concentration 1*10^-5, that means pOH = 5. So pH = 9. Then you compare pH to the pKa of the indicators. If the pH is higher than the pKa, than the indicator is deprotonated.


For 256, you have NH4+ ion. This is an acidic solution. However, ammonium is a weak acid. Therefore pH is below 7. Since it is not a strong acid, you know pH is above 2. That is why 4-6 would be the best answer.

Hope this helps. Keep up the great work!

 

[blue222]

Thank you @orgoman22 !

 

[DAT Destroyer]

Thank you @orgoman22 !

Always glad to help!