# Finding pH

1. Feb 8, 2006

### Chromium

I'm being asked to find the pH of a 0.050 M solution of NaCN.
I figured out the expression to be x^2/ 0.050 -x, now all i need is a value for k, however I'm not given that value. I know the pH depends on the molarity of the hydrogen ion at equilibrium, however to find that I need to find x, and in order to find x I need a value for k. I would appreciate any help, thanks.

2. Feb 8, 2006

### Gokul43201

Staff Emeritus
The expression for what ?
where 'x' is ...?

3. Feb 8, 2006

### Haxx0rm4ster

He is talking about the equilibrium constant equation.
Btw, i just finished coding a little simple program on finding the equilibrium constant.
Just saves a few keystroked on the annoying calculator :) .

water dissociation const.
http://en.wikipedia.org/wiki/Dissociation_constant

Last edited: Feb 9, 2006
4. Feb 9, 2006

### Staff: Mentor

pKa for HCN is 9.31.

Best,
Borek

Last edited by a moderator: Aug 13, 2013
5. Feb 9, 2006

### GCT

yeah, you should have been given the K value

6. Feb 9, 2006

### Chromium

I didn't check my book very well, and so later I found a chart that had the Kb value, so I could just used that to divide 10^-14 to find Ka, since Kb x Ka = Kw (10^-14). Once that's done I could have just used that value, plug it into the equation, solve for x, if it passes the 5% test then take the -log of x and had the pH.

7. Feb 9, 2006

### Chromium

I have no idea what pKa is... something with pressure? Thanks anyway though Borek. I got some help earlier, and was given the Ka constant, which was 6.2 x 10^-10.

8. Feb 9, 2006

### ksinclair13

There is a very good chance that all of this is wrong. I am basically just trying this problem out for myself as a challenge, for I don't know what pKa is either. I'm guessing...

$$pKa = - log Ka$$

$$9.31 = - log Ka$$

$$10^{-9.31} = Ka$$

$$Ka = 4.9*10^{-10}$$

$$4.9*10^{-10} = \frac{x^{2}}{.050 - x}$$

$$x^{2} + 4.9*10^{-10}x - 2.5*10^{-11}$$

$$x = 5.0*10^{-6}$$

$$5.0*10^{-6} = [HCN] = [H^{+}]$$

$$pH = - log (5.0*10^{-6})$$

$$pH = 5.30$$

Once again, I know that I am most likely wrong. I assumed that pKa = - log Ka. I also assumed that [HCN] = [H+]. I haven't learned about that yet; we just started acids and bases earlier this week in my class.

I'm interested to see how this problem is really meant to be worked out .

Last edited: Feb 9, 2006
9. Feb 9, 2006

### Chromium

Sorry, that was actually the Kb constant...so when you divide it out you get 1.6 x 10^-5 as Ka.

10. Feb 10, 2006

### aalmighty

pKa = -log(Ka)

11. Feb 10, 2006

### ksinclair13

So my assumption was correct . So did I do the problem correctly after all?

12. Feb 10, 2006

### Staff: Mentor

You guessed OK :)

Write the reaction equation - you don't start from HCN, you start with CN- - CN- will react with water to produce OH-. You are close, but you missed ;)

Best,
Borek

Last edited by a moderator: Aug 13, 2013