# GC questioin from 2010 Destroyer

#### i4everpaki

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??

10+ Year Member
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

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#### 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?