Question

calculate the pH of a solution that is made by combining 55 mL of 0.050 M HF with 125 mL of 0.10 M NaF. use an ICE table.

Answer 1

millimoles of HF = 55 x 0.050 = 2.75

millimoles of NaF = 125 x 0.10 = 12.5

HF -----------------------------------> H+   +   F-

2.75                                           0          12.5 --------------------> I

-x                                            +x           + x   ----------------------> C

2.75 -x                                   x           12.5 + x --------------------> E

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

6.3 x 10^-4 = (x)(12.5 + x)/(2.75 -x)

x = 1.39 x 10^-4

[x] = [H+]= 1.39 x 10^-4 M

pH = -log [H+]

pH = -log ( 1.39 x 10^-4 )

pH = 3.86