Destroyer gen chem Question 75

[dilinim]

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

I understand everything after you get the reaction.

The destroyer says "this problem shows how to calculate the pH of a salt. If you see Na, K, Li, or NH4, it is a salt. start the problem by removing the spectator ion and they get the reaction:
CN- + H20 --> HCN + OH-

How did they get this???? Did they just remove Na? And where did the H2O come from??

thank you in advance!
😳
I suck at gen chem sigh 🙁🙁🙁🙁

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[mario0923]

seriously I don't bother to "understand" this
I just memorized from Chad that if you know the concentration of HA and Ka
[H+]=sqrt((HA)*(Ka)) or [OH]=sqrt((A-)*(Kb))

if you want to know how he got it, he used the ice table.
HA -> H+ + A-

and for this question: the pH of salt is the hydrolysis of salt. in other words it is the reaction of the salt with water to break down
HA+H2O-> H3O+ + A-
or
A-+H20-> HA + OH-

so CN- + H20 -> HCN + OH-
you do the ice table
i 0.04 n/a 0 0
c -x n/a +x +x
e (0.04-x) n/a x x

so Keq=x^2/(0.04-x)
where you can neglect x in the denominator
Keq=x^2/0.04
so x=sqrt((0.04)*(Keq))
Keq here is the Ka that you are given
I just did the ice table to show you that the equation works

Hope it helped you 🙂

 

[jhc83]

Agreed, just know that it's a salt and plug in the information into the equation that Chad told you to memorize! Quick and easy. 🙂

 

[dilinim]

How do you use Chad's equation with this problem??? What exactly do you plug in and how do you end up with pH?

and how would you do the sqrt without a calculator?

THank youu and sorry I dont understand :/

 

[mario0923]

in this problem you have a salt NaCN which is Na+ CN- and you are given Ka of HCN
as I told you the pH of salt is the hydrolysis of salt which is
A-+H2O->HA+OH-
for this problem CN-+H2O->HCN+OH-
you do the ICE table as I have shown above where initial concentration of CN- is given initial concentration of H2O(l) is n/a initial concentration of HCN and OH- are zero (if you don't know how to do ice table I recommend you to study that first)
then you will get Keq=x^2([CN-]-x) where x is small enough to neglect so Keq=x^/[CN-]
Keq here is Ka since we are in equilibrium now
solve for x, x=sqrt((Ka)*([CN-]) which is same as the Chad's equation as I have shown above
if you got x, you can get the pH by the equation pH=-log([H+])

if you don't know how to do sqrt....well...
DAT will not give you anything hard that you can't do without caculator
for example if you have sqrt(4*10^-8) then you will have 2*10^-4 since sqrt of 4 is 2
sqrt of 10^-8 is 10^-4. this is simple high school algebra using powers. remember sqrt is like 1/2 power.

and if you don't know how to do logs...log of 10 to the power of something is just "something"
like log(10^9)=9
-log(10^-9)=9

dilinim, if you have a weak background on chemistry I recommend you Chad and destroyer 🙂

 

[dilinim]

Ok thanks! i think i get it now. It's just using Chad's equation does not work for this problem the way you put it.

Like: x=sqrt((Ka)*([CN-]) Does not work. I believe you are supposed to use Kb and not Ka in this example in order to get OH concentration first? Since OH is what is given off and it is a base we are using?

then get the pOH and subtract by 14 to get pH

 

[Miley13]

Ok thanks! i think i get it now. It's just using Chad's equation does not work for this problem the way you put it.

Like: x=sqrt((Ka)*([CN-]) Does not work. I believe you are supposed to use Kb and not Ka in this example in order to get OH concentration first? Since OH is what is given off and it is a base we are using?

then get the pOH and subtract by 14 to get pH

NaCN is a weak base (you know this by using Chad's list of negligible cations and anions; Na is negligible because it is a group 1 and CN- is not one of the negligible anions and therefore NaCN is basic).

This is a trick you definitely have to look out for because doing it using Ka will definitely be one of the multiple choice answers.

Use Kw = 1 x 10^-14 and Kw = Kb x Ka to solve for Kb. Then use chads equation. Hope this helps!