What must be the ratio of F- concentration to HF to make a buffer with a pH of 4? pKa = 3.2
a. 0.16
b. 2.4
c. 6.3
d.8.1
pH = 4
[H] =10-pH
= 10-4 = 1.0x10-4
Also;
[H] "=\\dfrac{nHF x Ka}{nF-}" Since pKa = -LogKa
Ka = 10-pKa = 10-3.2
1.0x10-4 "=" "\\dfrac{nHF x 10^-3.2}{nF^-}"
"\\dfrac{nHF}{nF^-}" = 0.158 = 0.16
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