[PJB]

Hi all,
 I'm starting to feel like I am probably the most chemistry-challenged student on the planet at the moment, so that means my questions will be easy ones to all of you!

At 25C, what is the PH of a 1.75 M solution of sodium cyanide?  For HCN, Ka = 4.0 x 10^-10.

My idea for the equation would be:  HCN + H2O = H3O + CN
Then I would say that 4.0 x 10^-10 = x^2 / 1.75.  I've tried the easy method of taking the square root of ka(1.75) and I've also tried the quadratic equation but nothing seems to come up with any of the answers listed in the multiple choices!  I'm not sure where I am going wrong.

Thanks in advance for the *delete me*
Pam

[Demotivator]

Don't confuse hydrolysis of a salt like CN- with acid dissociation like HCN.

Hydrolysis:
CN- + H2O -> HCN + OH-   Kh
solve for OH- , then PH

Kh = Kw/Ka

proof
H2O = H+ + OH-  Kw
H+ + CN- = HCN   1/Ka
therefore, net equation
CN- + H2O -> HCN + OH-   Kw/Ka

[jdurg]

Alright, this is a simple I.C.E. type of equation.  (I.C.E. means setting up a little table where you list the Initial, Change, and Equillibrium concentrations of all species involved in the reaction).  HCN is a weak acid, so when a cyanide salt is placed in water the solution becomes basic in nature.  (Because the CN- ions will grab an H+ from water to form HCN and leave OH- ions in solution).  Therefore, we will need to convert the Ka of HCN into the Kb of CN- to do the equation properly.  (Remember that Kb = Kw/Ka).  Since you know the Ka of HCN, you can find the Kb of CN- by using (1.0x10^-14)/(4.0x10^-10).  This will give you the Kb of CN-.

Now the equation you will need to use is the following:

Kb(CN-) = ([HCN][OH-])/[CN-]

Since you know the initial concentration of CN-, you can figure out the equillibrium concentrations of OH- and HCN.  

Initial:
HCN = 0
OH- = 0
CN- = 1.75 M

Change:
HCN = x
OH- = x
CN- = -x

Equillibrium:
HCN = x
OH- = x
CN- = 1.75 - x

Using the equation written out above, you solve for x.  Once you get the value of x, you'll then have the value of the OH- ion concentration at equillibrium.  If you take the -log of that concentration, you will have yourself the pOH of the solution.

Since pOH + pH = 14, you can quite easily determine the pH of the solution.

Let me know if you need further help with this.  

[Borek]

Don't confuse hydrolysis of a salt like CN- with acid dissociation like HCN.

I have a feeling that your answer is - unvoluntarily - confusing

While the effect of hydrolysis is exactly opposite to that of dissociation, both processes are described by the same equlibrium constant (that happens to be called dissociation constant) and Kw. That's the underlying idea of the Bronsted Lowry theory.

Treating hydrolysis and dissociation separately is dangerous, as it leads to confusion - do we have to concentrate on dissociation or hydrolysis? The only proper answer is 'we have to calculate equlibrium'. That is especially important when dealing with multiprotic acids.

Kh = Kw/Ka

Kh is nothing else than Kb.

[Demotivator]

Treating hydrolysis and dissociation separately is dangerous, as it leads to confusion - do we have to concentrate on dissociation or hydrolysis? The only proper answer is 'we have to calculate equlibrium'. That is especially important when dealing with multiprotic acids.

For this problem, concentrate on hydrolysis since it's equilibrium constant  dominates, making the approximation valid. Otherwise, the math gets complicated with simultaneous equations.

[Borek]

For this problem, concentrate on hydrolysis since it's equilibrium constant  dominates, making the approximation valid. Otherwise, the math gets complicated with simultaneous equations.

We agree that hydrolysis dominates. That's a pointer on how to do the calculations. But there is no separate hydrolysis equilibrium constant. Or rather it is an entity that should be removed with Ockham razor

[GCT]

Hi all,
 I'm starting to feel like I am probably the most chemistry-challenged student on the planet at the moment, so that means my questions will be easy ones to all of you!

At 25C, what is the PH of a 1.75 M solution of sodium cyanide?  For HCN, Ka = 4.0 x 10^-10.

My idea for the equation would be:  HCN + H2O = H3O + CN
Then I would say that 4.0 x 10^-10 = x^2 / 1.75.  I've tried the easy method of taking the square root of ka(1.75) and I've also tried the quadratic equation but nothing seems to come up with any of the answers listed in the multiple choices!  I'm not sure where I am going wrong.

Thanks in advance for the *delete me*
Pam

KaxKb=K water=1.00x10^-14

find Kb

Kb=[0H-][HCN]/[CN-]

=x^2/[initial concetration CN - x]

solve for x

p0H=-log[0H]