Question

Consider the titration of 42 mL of 0.111 MNaOH with 0.0803 MHCl. Calculate the pH after the addition of each of the following volumes of acid. 6.8mL, 42mL, .10L

Answer 1

HCl is strong acid and sissociates completely to give H+ and Cl-

NaOH is strong base and dissciaties completely to give Na+ and OH-

Initial NaoH moles = M x V = 0.111 x ( 42/1000) = 0.004662

when 6.8 ml i.e 0.0068 L HCl added the HCl moles = M x V =0.0803 x 0.0068 =0.000546

now 1HCl reacts with 1NaoH

excess NaOH after reacting with HCl = 0.00462-0.000546 = 0.004074

total vol = 42+6.8 = 48.8 ml = 0.0488L , [OH-] = moles/vol = 0.004074 /0.0488 =0.08348

pOH = -log [OH-] = -log ( 0.08348) = 1 , pH = 14-1 = 13

when 42 ml HCl added i.e 0.042L , HCl moles = M x V = 0.0803 x 0.042 = 0.0033726

excess NaOH moles = 0.004662-0.003376 = 0.001286

total vol of solution = 42+42 = 84 ml = 0.084L

[OH-]= ( 0.001286/0.084) =0.015

pOH = -log [OH-] = -log ( 0.015) = 1.8

pH = 14-1.8 = 12.2

when 0.1 L HCl added moles of HCl = 0.0803 x 0.1 = 0.00803

now exces HCl after neutralising NaoH= 0.00803-0.004662 = 0.003368

vol of solution = 0.1+0.042= 0.142 L

[H+] = ( 0.003368/0.142) = 0.02372

pH = -log ( 0.02372) = 1.6