Question

Determine the pH of an HF solution of each of the following concentrations. a. .290 M b. 5.20 x 10^-2 M c. 2.10 x 10^-2 M d. in which cases can you not make simplifying assumption that x i small

Answer 1

HF----> H+ F-

Ka for HF= [H+] [F-]/[HF] =3.39*10^-4

let x= drop in conc to reach equilibrium

at Equilibrium [HF] =-.29-x [H+] =[F-] =x

Ka= x²/(0.29-x)= 3.39*10^-4

since extent of dissociation is small, 0.29-x can be approximated to 0.29

x²/0.29= 3.39*10^-4, x=0.0099 this is much much less than 0.29M

pH= -log (0.0099)=2.00

b) for 5.2*10^-2M HF, writing expressions for ka as done above,

x^2/(5.2*10^-2-x)= 3.39*10^-4, using solver of excel x=0.0042, pH=2.376 ( x can't be ignored in this case)

c. for 2.1*10-2 M HF, x²/(2.1*10^-2-x)= 3.39*10^-4, x= 0.002668, pH= -log(0.002668)= 2.57 ( x can''t be ignored in this case)