ph problem 2222?

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The K(a) of HF at 25.0 deg. C is 6.8 x 10^-4. What is the pH of a 0.35 M aqueous solution of HF?

show work please

2007-04-21 10:09:07 · 2 answers · asked by Zywiec 2 in Science & Mathematics ➔ Chemistry

Ka=[H+]^2/[HF]

[H+]^2=6.8x10^-4 x 0.35

pH= -log (square root of 2.38x10^-4)

2007-04-21 11:06:09 · answer #1 · answered by cyprus1988 2 · 0⤊ 0⤋

Take the negative log of (root Ka x 0.35).

2007-04-21 17:42:50 · answer #2 · answered by Gervald F 7 · 0⤊ 0⤋