The pH of a 0.25 M aqueous solution of hydrofluoric acid, HF, at 25.0 °C is 2.03. What is the value of Ka for HF?
6.0 × 10-5
3.5 × 10-4
1.1 × 10-9
2.0 × 10-9
none of the above

Question

The pH of a 0.25 M aqueous solution of hydrofluoric acid, HF, at 25.0 °C is 2.03. What is the value of Ka for HF?
	 6.0 × 10-5
	 3.5 × 10-4
	 1.1 × 10-9
	 2.0 × 10-9
	 none of the above

anonymous · Answer

$$pH = pK_a-\log_{10}[HF]$$plug in the numbers

anonymous · Answer

$$2.03=pK_a-\log_{10}(0.25)\
pK_a=?
$$

anonymous · Answer

2.28?

anonymous · Answer

OH wait! I think I did that wrong.

anonymous · Answer

Okay, so i'm taking pH then adding the log of (.25) correct? When you put log10 is that different from the log button on my calc that just says log?

anonymous · Answer

log = $\log_{10}$

anonymous · Answer

you should get pKa = 1.428

anonymous · Answer

gotcha, so the answer i get when i add 2.03 and log(.25) is 1.4279 for pKa

anonymous · Answer

then I have to change that to Ka by inverse log, right?

anonymous · Answer

Nevermind, thanks tho. I guessed and got it right.

anonymous · Answer

$$
pH=-\log_{10}[H^+]\
[H^+]=10^{-2.03}=0.00933\
[F^-]=[H^+]=0.00933\
[HF]=0.25-0.00933\
\;\
K_a=\frac{[H^+][F^-]}{[HF]}\
K_a=\frac{0.00933\times0.00933}{0.25-0.00933}\
\Large\boxed{K_a=3.6\times10^{-4}}
$$