# Experiment's ph, simple i think

1. Mar 12, 2006

### Pengwuino

In an experiment we did, we simply added 5ml of 0.1ml HCl into 20ml of a .5M solution of an unknown weak base. I'm not sure how we calculate the equilibrium constant though

We know the pH is 5.855 which means the pOH is 10^-(14-5.855) = 7.161 x 10^-9.

Now i'm a bit confused. Is the [HB+] concentration the same as the [OH-] concentration and is the equilibrium equation this?

$$Keq = \frac{{[HB][OH]}}{{}}$$

2. Mar 18, 2006

### siddharth

I think some more information needs to be known. For instance, is it a mono-acidic base (like NaOH)?

What you need to find is which is in excess (ie, at equilibrium, what is the concentration of each species? Is there an excess of the acid, the base or did they react completely?)

The concentration of OH- will be 10^-(14-5.855) = 7.161 x 10^-9 not pOH.

First of all, how did you get that equilibrium equation?
As I said earlier, you need to know what the concentration of the species at equilibrium are before you can find the equilibrium relation

Last edited: Mar 18, 2006
3. Mar 18, 2006

### GCT

siddharth gave some very good points, however, the premise here is the equation pKa=pH at the half equivalence point as far as this lab is concerned. Both of you should note that, the mixing of the acid and base would result in half of the base titrated.

4. Mar 18, 2006

### Staff: Mentor

There is something wrong with the question - there is no base that will give acidic solution when partially neutralized.

Not 50% but 5% of the base was neutralized, so pKa is not equal to pH.

Pengwuino - take a look at this question: calculation of pKa of a weak acid. It describes similar case, so should be helpfull.

5. Mar 18, 2006

### siddharth

Why is that? If it is a mono-acidic base like NaOH, then initially there are 10 mmol of NaOH and 0.5 mmol of HCl. So only 0.5 mmol of the base will react.

Yeah, since there is an excess of the base, I would have thought that the resulting solution will be basic.

6. Mar 18, 2006

### Pengwuino

I figured the equation should be...

$$HA_{aq} + B_{aq} \leftrightarrow HB_{aq} + OH_{aq}$$

HA being the HCl
B being the weak base
HB being the salt
OH being hydroxide
H2O being some strange liquid :P

Then the equlibrium equation should be (after reducing)

$$\frac{{[HB][OH]}}{{}}$$

Since all the acid was neutralized (or at least i suspect it should have)

Last edited: Mar 18, 2006
7. Mar 19, 2006

### GCT

yeah, the problem's messed up to begin with, we've got nothing to work with here really so it's somewhat senseless to continue the discussion, so Penguino should restate the problem,

that was stupidity on my part, I assumed that since the experiment involved a simple step to calculate the Ka, that the case here was that of half equilvalence.

8. Mar 19, 2006

### Pengwuino

Well the problem was very straight forward. We had a solution of HCl and a solution of an "unknown weak base".

I Just realized i should have put "5ml of 0.1M HCl". That was basically what we had to work with and i just assume it. I also skipped a step and said my [OH] was my pOH without clarifying that i was converting to [OH] already.

9. Mar 19, 2006

### GCT

if it were the simple case where we're dealing with a half-equivalence point problem, then pKa=pH, but from the problem, it doesn't seem that way.

in reference to your original post, the hydroxide term will equal the acid term, before you add the acid. You'll need to make some subsequent rearrangements and then work backwards to solve for Kb. Note that they give you the pH (after adding the acid?), you can use this to solve for the hydroxide term. Let's see what you can do.

10. Mar 20, 2006

### GCT

$$Kb=[HA][OH-]/[HA]=[x][x]/[.5M-x]$$ and subsequently

$$Kb=[.0005M+x][10^{-8.145}]/[.5M-x-.0005M]$$ assuming we're dealing with the buffer region (which may or may not be the case here).

11. Mar 20, 2006

### GCT

So you've got two equations to work with

12. Mar 20, 2006

### Staff: Mentor

$$Kb=[HA][OH^-]/[HA]=[OH^-]$$ ?

13. Mar 21, 2006

### GCT

yeah, thanks for pointing that out

$$Kb=[HA][OH-]/[A-]$$