[tjpj77]

Find pH if 3 mL 0.1M HCl is added to 97 mL of pure H2O at pH 7.0.

(3mL/100mL)(0.1 mol HCl/1L) = 3 x 10^-3 mol HCl/L
3 - log 3 = 2.5
pH = 2.5
right?

[Dan]

Find pH if 3 mL 0.1M HCl is added to 97 mL of pure H2O at pH 7.0.

(3mL/100mL)(0.1 mol HCl/1L) = 3 x 10^-3 mol HCl/L

OK

3 - log 3 = 2.5

What are you doing here? Where does this come from?

[tjpj77]

after you have the M of a strong acid or base, you can just do the quick math via that trick: bring down and cross over the power and then subtract from that the log of what was the first number.

[Borek]

Hardly a trick, just using log properties:

log(3×10^-3) = log(3) + log(10^-3) = ~0.5 - 3

[Dan]

Ah OK. I agree with 2.5, I was just wondering where that calculation came from. I've never bothered to expand the log before calculating for questions like this.

[Borek]

For me it was obvious.

Most likely that shows the difference between me and the calculator generation 

[Dan]

I was thinking similarly