Given 100.0 mL of a buffered solutions which is 0.30 M #HF# and 0.30 M #NaF# and the Ka of #HF# to be 7.2 x 10^4, how do you find the pH of this buffer?

Question

1761 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-20T12:05:17

I assume that the value of #sf(K_a)# is #sf(7.2xx10^(-4)color(white)(x)"mol/l")#.

#sf(pH=3.14)#

#sf(HF)# dissociates:

#sf(HFrightleftharpoonsH^(+)+F^(-))#

For which:

#sf(K_a=([H^+][F^-])/([HF])#

These are equilibrium concentrations.

Rearranging gives:

#sf([H^+]=K_axx([HF])/([F^(-)])#

Assuming we have been given equilibrium concentrations these will be the same so:

#sf([H^+]=K_(a)xx(cancel([HF]))/(cancel([F^(-)]))#

#:.##sf([H^(+)]=7.2xx10^(-4)color(white)(x)"mol/l")#

#sf(pH=-log[H^+]=-log(7.2xx10^(-4))=3.14)#