Question

Part A
How many moles of HF (Ka=6.8

Answer 1

we know that HF is a weak acid and will undergo dissociation as

                   HF    -->         H+ + F-

inital            a                         0            0

at equili       a-x         x               x

ka = [H+][F-] / [HF]

Ka = x^2 /a -x

x<<<1 for weak acid

so Ka = x^2 / a = 6.8X 10^-4

now x = [H+]

pH = 3

so [H+] = antilog of -3 = 0.001

so

0.001 X 0.001/ a = 6.8X 10^-4

or a = 0.001 X 0.001 / 6.8 X 10^-4 = 1.47 X 10^-3M or moles / L

now we want to know moles in 0.15 L

which will be = 1.47 X 10^-3 X 0.15 = 2.205 X 10^-4 moles

Or without assuming x<<1

Ka = x^2 / a-x = 6.8X 10^-4

now x = [H+]

pH = 3

so [H+] = antilog of -3 = 0.001

so

0.001 X 0.001/ a-0.001 = 6.8X 10^-4

Or 10^-6 = 6.8 X 10^-4 a