gen chem -- I can't do math

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gooperwooper

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for example, find pH of a 0.03M NaCN solution. The Ka of HCN is 6x10^-10.

How would I even begin to do this without a calculator?
 
I believe you convert HCN to Kb then do sqroot(Kb*[-CN]) to find pOH. then 14 - ans
 
This is my biggest problem. I can find PH without a calculator for strong acids, but these weak acid/base calculations without calc is killing me. Even after Chads videos, I'm still lost :hungover:
 
Here is my take on this kind of problem without a calculator:

First, you need to memorize these for all pH calculations

A. pH or OH of 2 x 10^-n ~ (n-1). 7
B. pH or OH of 3 x 10 ^-n ~ (n-1).5
C. pH or OH of 8 x 10^ -n ~ (n-1).1

Next, find out what is the Sqrt (Kb x [CN]) is roughly equals to. Example, lets say [OH] = Sqrt (Kb x [CN]) = 5 x 10^-8. From here you can calculate the pOH is roughly between 8 and 7.

In order to estimate it closer, you use the three values listed above. In this case, 5 x 10^-8 is closer to 3 x 10^-8. So you can deduce that pOH is around 7.3-7.4.
 
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I found that chads way of estimating logs didn't work for me. I got lost in his explanation of what to memorize and my lack of ability to visualize log graphs. This guy does a much better job at explaining it.


For example if i wanted to know the pH of 5x10^-8 i would take the power of the 10, so 8, and subtract .7, the log of the base number, based on the table in the video to get a pH 7.3 . (in the end you would only have to memorize the table below, which comes with practice) As for solving square roots and other arthritic, its much easier to write everything in exponential form and cancel as many zeros out. This is what works best for me anyway*

pH table
#, Log
1, 0
2, .3
3, .48
4, .6
5, .7
6, .78
7, .85
8, .90
9, .95
10, 1
 
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