GC questioin from 2010 Destroyer

[i4everpaki]

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Find the pH of a 0.040 M NaCN solution. Given: Ka of HCN is 5.0 x 10^-10

a- 11
b- 8.7
c- 5.3
d- 12
e- 3.4

this problem was really difficulty for me, and i could not understand how to solve it until i looked at the ans key. there were many steps and esp calculating with log. the ans is choice a 11. is there an easy way to solve this probably, some neat trick or anything? how would you solve this prob??

 

[UndergradGuy7]

Na+ is a spectator ion. First write the reaction. CN- + H2O -> HCN + OH-
We know the concentration of CN- as 0.040 M. Do an ice table or memorize the equation for this kind of problem.
If the solution reacts then CN- concentration will be 0.040 M - x (some of it will turn into products but not all. The amount that turns into products is x). And HCN and OH- will be made as "+x". Also notice this is true for this reaction because the coefficients are 1 for everything. If they were different the setup would change a little.

So do the typical equation products/reactants = K you get [(x HCN)(x OH-)]/(0.040 - x CN-) = K

Here they give you Ka the acid constant, but the equation in the forward direction is for CN-, a base. So we need to convert Ka to Kb. You do this by (1.0E-14)/(Ka) = 2.0 E -5

[(x HCN)(x OH-)]/(0.040 - x CN-) = 2.0 E -5

Solve and you will have to use the quadratic equation. If you don't want to, sometimes you can say and x in the denominator is very small compared to 0.040 M and we can ignore it so we can simplify the equation to [(x)(x)]/(0.040M) = 2.0 E-5
We can do this because CN- is weak and should not react that much. You can also do (HA or A-)/K and if it is >100 the approximation is OK.
x^2 = 8.0E-5 take square root and you get x=8.94E-4 M
x is the concentration of OH so if we took -log(x) we would get pOH = 3.04. The problem asks for pH and pH + pOH = 14. So we can do 14-3.04 = 10.96 = pH

 

[i4everpaki]

how are you able to do logs quickly in your head? i am so use to using a calculator... is there a neat trick or some way to do logs without using the calculator?

 

[UndergradGuy7]

The kaplan book explains it well. What you do is first write the number in scientific notation like 2.0 x 10^-5
There is a log rule where log(2.0 x 10^-5) can be rewritten as -5+log(2.0) since log(10^x) = x. Since for our equation we are doing the -log our problem is 5-log n. So now we know 5 - log👎. So the answer has to be around 5. But we also know that log can be either 0 if we do log(1) and 1 if we do log(10). So the log👎 term will be a number between 0 and 1. The bigger the number the closer to 1 it is and the smaller the number the closer to 0 it is.

So in the example above, 5-log(2.0). 2.0 is close to 0. 2/10 is 0.2. So we can estimate 5-0.2=4.8. If you did the answer on your calculator the answer is 4.7 so we are pretty close.

 

[i4everpaki]

Thank you so much, this makes it so simple and clear. If you have any other tricks please share. Thanks alot