Question

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

Answer 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