• Practice your interview with the new SDN AI Interview Coach. Choose a school, answer by voice or typing, and receive a personalized feedback report. Available now to all SDN members. Try the AI Interview Coach.

Destoryer 2016 # 226 GC

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

Fancy312

Full Member
10+ Year Member
Advertisement - Members don't see this ad
Soooo I thought I had the hang of this problem down till I realized i got it wrong and still confused on the log portion of the problem:::

226. Consider a sample of 100 mL of pure water @ 25 degrees celsius. If the hydronium ion concentration was tripled, what would the resulting pH be?

A. 6.5
B.4.8
C 7.0
D. 3.5
E. 2.0

The solutions explains how tripling the H3O+ results in::
3(1 x 10^-7)= 3 x 10^-7 M
pH= -log 3 x 10 ^-7 which roughly equals around 6.5

WHERE DID 6.5 COME FROM?! I'm so bad at logs and estimating them and even went to that SDN thread that talked about how to do logs but its still frustrating :'/
 
The -log(1 x 10^-7) is equal to 7. The -log of (1 x 10^-6) is 6.

3 x 10^-7 is somewhere in between (1 x 10^-7) and (1 x 10-6), right? Because if you had 9.99 x 10^-7 and added 0.0000000001, you'd get 10^-6. Realizing that 10^-6 is actually larger than 10^-7 is the trick to estimating logs correctly.

So the correct answer must be between 6 and 7, so 6.5 is an acceptable answer choice.
 
Soooo I thought I had the hang of this problem down till I realized i got it wrong and still confused on the log portion of the problem:::

226. Consider a sample of 100 mL of pure water @ 25 degrees celsius. If the hydronium ion concentration was tripled, what would the resulting pH be?

A. 6.5
B.4.8
C 7.0
D. 3.5
E. 2.0

The solutions explains how tripling the H3O+ results in::
3(1 x 10^-7)= 3 x 10^-7 M
pH= -log 3 x 10 ^-7 which roughly equals around 6.5

WHERE DID 6.5 COME FROM?! I'm so bad at logs and estimating them and even went to that SDN thread that talked about how to do logs but its still frustrating :'/

When you reach the DESTROYER buffer question on question 123 solution, I give you a lot of practice in estimating logs. This is a simple skill that you must master. If you took the negative log of 1 x 10 exp -4.....you get 4. If you are at a value, 2 x 10 exp -4...you need to Decrease your estimate a tad bit.thus, 3.8 would be reasonable. I give several great examples for you to practice on...try them.

Hope this helps

Dr. Jim Romano