[guybrush]

I have been asked to work out the pH of a weak acid.

I have been given the mol of it (0.5 mol) and the volume (2.0dm 3). The concentration is 0.25 mol dm -3.

Next i have the Ka which is 1.4 x 10 -3. pKa = 0.0014 -log which equals 2.85.

I know the pH solution is: 1/2 x pKa - 0.5 (-log concentration)

Going by this i have:

= 0.5 x 2.85 - 0.5 (-log 0.25 mol dm -3)

= 1.425 - 0.5 ( - 0.6)

= 1.425 + 0.3

pH = 1.73

Am i way off the mark or am i sort of there?

[Borek]

No idea what you are doing. Please elaborate using standard math notation

pKa = 0.0014 -log which equals 2.85

means nothing.

[guybrush]

Sorry, to get to 2.85 i used the equation is: pKa = -log (Ka).

Ka = 1.4 x 10 to the power -3.

[Borek]

Explain what you did later.

I suppose you have used some formula, I would have to analyze whole post to guess what it was.

[guybrush]

Thanks for the reply, here goes:

To calculate the pH i have used the formula: pH = 1/2 (0.5) x pKa (2.85) - 1/2 (0.5) x (-log concentration). The concentration is 0.25

To calculate the concentration i used the formula: concentration = amount (0.5 mol) / volume (2.0dm to power of 3)

The concentration came to (0.25).

My equation now reads

ph = 0.5 x 2.85 = 1.425   -   0.5 x (0.6 this is the -log concetration result) = 0.3

ph = 1.425 - 0.3 = 1.125. Sorry got my - and + mixed up.

[guybrush]

Meant to add, to get the pKa i used: 1.4 x 10 to the power -3 and pressed -log which gave me the 2.85

[Borek]

So basically that was [H⁺] = sqrt(Ka*C) - do you know when this formula can be used?

http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base

[guybrush]

Thanks for the reply.

Using the equation you said gives the same answer (1.73 originally) as mine but without going the long way round.

Using H+ = Ka*C sqrt gives me 0.0187. If i put this into the equation below like so:

pH = -log (H+)

pH = -log (0.0187)

pH = 1.73

I also used this equation 8.10 of that link you sent and it gave me the same.


It can be used when H+ is less than 5% of the Concentration.

Thankyou for your help.

[Borek]

It can be used when H+ is less than 5% of the Concentration.

Have you checked if the condition is meet?

[guybrush]

5% of the concentration is: 0.0125

H+ answer = 0.0187

This means the condition has not been met so i would need to use a different equation ie the one below:

H+= -Ka + (Ka2 + 4Ka x Ca sqrt) / 2

H+ = -1.4 x 10 -3 + (1.96 x 10 -6 + 0.0056 x 0.25 sqrt) / 2

H+ = -1.4 x 10 -3 + (0.0374)/2

H+ = -1.4 x 10-3  + 0.0187

H+ = 0.0173

pH = -log H+ (0.0173)

pH = 1.76

[Borek]

You should not round down intermediate results. Otherwise you are on the right track.

[guybrush]

You should not round down intermediate results. Otherwise you are on the right track.

Think i made a calculation error by not added -Ka to 0.0187 before dividing by two. However worked it out fully now and i get 1.74, this is using the full answers in the intermediate results.

Thankyou for your guidance.