How do you calculate the pH of a weak acid?

2 Answers
Sep 17, 2015

Answer:

A handy expression to use is #pH=1/2(pK_a-loga)#

Explanation:

This is a quick and useful method if you are given the #pK_a# and concentration #a# of the weak acid.

I'll derive it using the usual ICE method:

#HX_((aq))rightleftharpoonsH_((aq))^(+)+X_((aq))^(-)#

#"Initial:"## " "#a#" "" "0" " " " " "0#

If #x# moles dissociate the equilibrium moles #rArr#

#(a-x) " " " " " " " " x " " " " " " x#

#K_a=(x^2)/(a-x)#

We assume that #x# is negligible compared to to #a# so this approximates to:

#K_a=(x^2)/(a)#

#(x^2)/(a)=K_a#

#x^2=K_axxa#

#x=(K_axxa)^((1)/(2))#

Taking -ve logs of both sides #rArr#

#-logx=-1/2log(K_a+a)#

#pH=1/2(pK_a-loga)#

Here's an example:

#pH# of a #0.1"M"# solution of ethanoic acid whose #pK_a=4.75# ?

#pH=1/2(4.75-(-1))#

#pH=5.75#

If you can remember it then fine. If not you need to use the ICE method described in the answer by @Dr Hayek

Answer:

#pH = -log[H_3O^+]#

Explanation:

By measuring the concentration of #H_3O^+# ions we get the pH using #pH = -log[H_3O^+]#

Consider the dissociation of the following weak acid HA with initial concentration of #1.0 M# and #K_a = 2.0 xx 10^-6#:

#" " " " " " " " " " "# #HA(aq) + H_2O(aq) -> H_3O^+(aq) + A^(-)(aq)#
#" " " " " " # Initial: #1.0 M " " " " " " " " " " " " 0 M " " " " " " 0M#
#" " " " " # Change: # - xM " " " " " " " " " " " +x M" " " " " +x M#
#" " " # Equilibrium: # (1.0- x)M " " " " " " " " "x M" " " " " " " " x M#

The expression of #K_a# can be written as follows:

#K_a = ([H_3O^+(aq)]*[A^(-)(aq)])/([HA(aq)])=(x*x)/(1.0-x)=2.0xx10^-6#

solve for #x# using calculator, you get

#x~~1.4xx10^-3 => pH = -log(1.4xx10^-3) = 2.895#

that can be rounded to #pH=2.9# (2 significant figures).

Here is a video which summarizes this technique which requires use of the quadratic formula. The values are different than Dr. Hayek's example, but the technique is the same.

Video from: Noel Pauller