If the #K_a# of a monoprotic weak acid is #3.4 x10^-6#, what is the pH of a 0.14 M solution of this acid?

1 Answer
Jan 28, 2017

Answer:

#sf(pH=3.15)#

Explanation:

Let #sf(HX)# be the weak acid:

#sf(HXrightleftharpoonsH^++X^-)#

For which:

#sf(K_a=([H^+][X^-])/([HX]))#

If #sf(K_a)# lies in the range #sf(10^(-4))# - #sf(10^(-10))# we can assume that the equilibrium concentrations used in the expression are a good enough approximation to the initial concentrations.

This applies here.

Rearranging and taking negative logs of both sides gives:

#sf(pH=1/2[pK_a-loga])#

#sf(a)# is the concentration of the acid.

#sf(pK_a=-logK_a=-log(3.4xx10^(-6))=5.46)#

#:.##sf(pH=1/2[5.46-(-0.854)]=3.15)#