I'm having trouble with chemistry - can anyone help, please?

[Christine S]

We're going over pH values in my chemistry class, I'm having a hard time figuring out if I'm doing this right. If you know how to do this will you tell me if I'm right & if I'm not will you help me figure out how to do it correctly? I can't get my brain wrapped around this!

What are the pH values of the following solutions?
A) 0.15 M HNO3
= log(0.15) = -0.82 = 0.82
B) 1.5 x 10^-3 M HNO3
= -log(10^-3) = 2.83
C) 0.15 M NaOH
= log(0.15) = -0.82 = 0.82
D) 1.5 x 10^-3 M NaOH
= -log(10^-3) = -2.83 = 2.83

I'm sure I'm doing this wrong! I feel like I'm leaving a bunch of stuff out, but my book doesn't explain it very well and I'm lost! If anyone can tell me what I'm missing (preferably in the form of steps to follow, etc. rather than just giving me the answers) I'd really appreciate it!

Thanks in advance!

 

[jacee]

You did the first two right, but the second two are solutions of a strong base, not a strong acid. There are two common ways to find the pH, I think this one is easier:

C) Since you know the concentration of strong base, you can find pOH, which is -log[OH-] (works just like pH)

pOH= -log(.15) = .82

The pOH and the pH always add to 14, so the pH of this solution is actually 14-.82= 13.18, which makes sense because basic solutions have high pH.

D) Do the same thing. the pOH is 2.83 (-log[OH-], remember?), so the pH is 14-2.83= 11.17


Hope this helps.

 

[jacee]

You did the first two right, but the second two are solutions of a strong base, not a strong acid. There are two common ways to find the pH, I think this one is easier:

C) Since you know the concentration of strong base, you can find pOH, which is -log[OH-] (works just like pH)

pOH= -log(.15) = .82

The pOH and the pH always add to 14, so the pH of this solution is actually 14-.82= 13.18, which makes sense because basic solutions have high pH.

D) Do the same thing. the pOH is 2.83 (-log[OH-], remember?), so the pH is 14-2.83= 11.17


Hope this helps.