shortcut for this pH problem?

[sw1sschez]

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Find the pH of a 0.040 M NaCN solution. Ka of HCN= 5 x 10^-10.

I thought about using the pH=sqrt of (Ka*[H+]) formula but it doesnt work.

 

[dynococus]

Find the pH of a 0.40 M NaCN solution. Ka of HCN= 5 x 10^-10.

I thought about using the pH=sqrt of (Ka*[H+]) formula but it doesnt work.

Is the answer pH=11? Here's how I got it:

The chemical equation is: NaCN + H2O --> HCN + OH-

Kb = [HCN][OH-]/[CN-], but you're given the Ka, not Kb, so convert Ka into Kb by using Kw/Ka = Kb. (1x10^-14)/(5x10^-10) = 2x10^-5 = Kb.

It's easier to find the pOH with Kb. Using Kb = [HCN][OH-]/[CN-], we know Kb and [CN-], so substituting we get 2x10^-5 = [HCN][OH-]/0.4M. Solving, we get [HCN][OH-]=8x10^-7. HCN and OH- will have the same concentration (1molHCN:1molOH-), so substitute: [OH]^2 = 8x10^-7. Estimate the square root and you should get about 9x10^-4=[OH-].

pOH = -log[OH-], so -log9x10^-4 is about 3 or 3.1=pOH. pH = 14-pOH so 14-3 = 11 is the answer.

 

[dennisDDS]

Find the pH of a 0.40 M NaCN solution. Ka of HCN= 5 x 10^-10.

I thought about using the pH=sqrt of (Ka*[H+]) formula but it doesnt work.

Dyno did it right too..
i think your main problem was the conversion of Ka..

Kakb = kw.
kakb = 10^-14
5x10^-10kb = 10^-14
kb =10^-14/5x10^-10 ==> 1/5 x 10^-4 or .20 x 10^-4

.40M X .20 = .08 X 10^-4

SQROOT [ 8 X 10^-6]
= something between 2-3 x 10^-3
= 3 - log (2.5) = 3 - .3 = 2.7

14 - 2.7 = 11.3

 

[sw1sschez]

dennisDDS: yup you are right. i had to do the conversion. but where did you get this from?

.40M X .20 = .08 X 10^-4




dynococcus: yes that is the correct answer.

thanks you 2. back to the drawing board. 👍

oh, I goofed with the original post it's actually .04M and not .40M

How do you guys square root 8 x 10^-7 in your head?

 

[dennisDDS]

dennisDDS: yup you are right. i had to do the conversion. but where did you get this from?

.40M X .20 = .08 X 10^-4




dynococcus: yes that is the correct answer.

thanks you 2. back to the drawing board. 👍

oh, I goofed with the original post it's actually .04M and not .40M

How do you guys square root 8 x 10^-7 in your head?

.40M X .20 = .08 X 10^-4

you know when do you .4(.2 x 10^-8) well i just spaced it out so it would be .08 x 10^-8


Its pretty easy.. once you learn the pattern..

for anything 10^-X If X is even then all you do is divide x by 2
so for instance if it was 10^-12 the square root of that is 10^-6..

as far as the number in front.. you should know the basic squares such as
2 x 2 = 4
3 x 3 = 9 etc..

so if you had a answer like 8
you know that 3 x 3 = 9
and 2 x 2 = 4

so 8 lies in between 4 - 9 so you can roughly guess that is going to be like. 2.6 - 2.7 since its closer to 9.

for logs you know how to do the 10^-x, whatever x is becomes a whole number for example -log(10^-4) = 4

and for the 2.7 just know that whatever you log something like log(3) it will be .1 - .2 higher than the number your logging for example..

log (2) would be like .3 - .4

so to sum -log (2.7 x 10^-4) would be

4 - log 2.7 or 4 - .3 = 3.7...

i hope that made sense..

 

[dynococus]

dennisDDS: yup you are right. i had to do the conversion. but where did you get this from?

.40M X .20 = .08 X 10^-4

dynococcus: yes that is the correct answer.

thanks you 2. back to the drawing board. 👍

oh, I goofed with the original post it's actually .04M and not .40M

How do you guys square root 8 x 10^-7 in your head?

I usually just multiply by a factor of 10 so that the number becomes easier to take the square root of. So 8x10^-7 turns into 80x10^-8. Taking the square root of 80 is roughly somewhere a little less than 9, so let's say 8.9. Then take the square root of 10^-8, which is 10^-4. So the square root of 8 x 10^-7 is estimated to be 8.9x10^-4 (basically 9x10^-4)