calculate the PH of a solution formed by mixing 85 ml of 0.13 M lactic acid with 95 ml of 0.15 M sodium lactate. (K HLaC =1.4 * 10^-4)

Ans:- 

this is a buffer soluton of lactice acid and its conjugate base 

          moles of salt = 95*0.15  

                               = 14.25 mili moles 

reaction of salt 

          NaC₃H₅O_{3 <--->} Na+    +  C₃H₅O_{3 -} 

               [C₃H₅O_3-]= 14.25 mili moles

          moles of acid = 85*0.13 

                                 = 11.05 mili moles 

Henderson Hasselbalch equation for a weak acid

                            pH = Pka  +  log ([A¯] / [HA])

so formula change to 

                          pH = Pka  +  log ([ C₃H₅O₃¯] / [C₃H₆O₃])

plug the value in formula, we get 

                         pH = - log KHlac   +  log 14.25/11.05

                         pH = - log 1.4 * 10^-4  +  log 14.25/11.05

use calculator to solve these values, we get 

                         pH = 3.85+ 0.11

                         pH =3.96

Answer = pH of solution is 3.96  

In this problem we have learned