Calculate #pH# of buffer solution with #0.1# #M# each of #CH_3COOH# & #CH_3COONa#. What will be the change in #pH# on addition of the following: ? (#pK_a# = #1.8 xx 10^-5#). The total volume of solution is #1# #L#.

(i) #0.01# #M# #HCl#
(ii) #0.01# #M# #NaOH#

1 Answer
Jun 22, 2017

Answer:

The buffer equation holds that.........

#pH=pK_a+log_10{([A^-])/([HA])}#

Explanation:

#pH=pK_a+log_10{([A^-])/([HA])}#

Where #HA# is the parent acid; and the #[A^-]# is its conjugate. For the derivation of this formula, which an undergraduate should be able to perform, see here.

And thus for this problem..........

#pH=4.76+log_10{([AcO^-])/([HOAc])}#

And note that #4.76=-log_10(1.8xx10^-5)#

Now the starting conditions specify that there are equal concentrations of acetate and free acid, and thus #pH=pK_a=4.76#. Why? Because #log_(10)1=0#, so given the equation, #[AcO^-]=[HOAc]#, and #([AcO^-])/([HOAc])=1#.

Anyway, work thru the problems, and post them back in this thread. The #pH# will be round about #4.76#, i.e. the #pK_a# of the parent acid.