Someone help me with this please

[prsndwg]

---

Members don't see this ad.

I am confused on Destroyer Gchem # 33 and the Q is:
Find PH of a 0.04 M NaCN soln. GIven: Ka of HCN is 5* 10^-10.

I keep getting 5.3 where the soln shows 11.

Can someone explains this plz?

Thanks

 

[GOLF]

yeah - the NaCN solution is going to react with h2o to form a strong base, NaOH. therefore the expression that would define this reaction would be Kb. The problem gives you Ka, and the fact that it is HCN just means it is the conjugate acid of CN-. So you have to recall what Kw is (Ka*Kb = 1E-14) and divide Kw by Ka to give you the Kb, which is the expression you need to find the pH of the NaCN solution. now if you set up the equation like regular, you should have 2.0E-4 = (x^2)/.04. The 0.04 comes from the concentration of the solution - you can neglect the amount of x subtracted from it, since it probably won't affect the pH that much anyway. solving for x^2 we get .08E-5 or 0.8E-6. Take the square root to find x (remember we're trying to find the [OH-] concentration) and we get 1E-3 (rough estimation - dunno that decimal exactly but it doesn't matter.) This gives you a pOH of 3, meaning we have a pH of 11.

might have an error or two in that explanation, because i found an error in my work in the middle, but it's right for the most part.

 

[prsndwg]

Thanks you for your response..
can you plz explain why cant I use this equation instead:

HCN + H2O--> H3O + CN-

yeah - the NaCN solution is going to react with h2o to form a strong base, NaOH. therefore the expression that would define this reaction would be Kb. The problem gives you Ka, and the fact that it is HCN just means it is the conjugate acid of CN-. So you have to recall what Kw is (Ka*Kb = 1E-14) and divide Kw by Ka to give you the Kb, which is the expression you need to find the pH of the NaCN solution. now if you set up the equation like regular, you should have 2.0E-4 = (x^2)/.04. The 0.04 comes from the concentration of the solution - you can neglect the amount of x subtracted from it, since it probably won't affect the pH that much anyway. solving for x^2 we get .08E-5 or 0.8E-6. Take the square root to find x (remember we're trying to find the [OH-] concentration) and we get 1E-3 (rough estimation - dunno that decimal exactly but it doesn't matter.) This gives you a pOH of 3, meaning we have a pH of 11.

might have an error or two in that explanation, because i found an error in my work in the middle, but it's right for the most part.

 

[Kneecoal]

Thanks you for your response..
can you plz explain why cant I use this equation instead:

HCN + H2O--> H3O + CN-

in the explanation it states that you're using NaCN. That's a salt. So in your equation, your reactant is the ion coming from the salt - NOT HCN, because that's not what you're given to start with. So once you get rid of the spectator ion, you'll see that:

CN- + H2O --> HCN + OH-

 

[mestmet31]

just think that any salt is a base (aside from your regular bases with an OH)...so if they give you a Ka..you will have to find the Kb..this is done by dividing the Ka by 1 * 10^-14...after you get your Kb you can solve for pOH and then subtract from 14

 

[prsndwg]

now I understand it .. thanks everyone