Question

Consider the titration of a 23.4 −mL sample of 0.125 M RbOH with 0.105 M HCl. Determine each quantity:

the pH after adding 4.1 mL of acid beyond the equivalence point

Express your answer using two decimal places.

Answer 1

RbOH is a strong base as well as HCl is a strong acid

To calculate the pH of the solution we have

As the concentration of H⁺ is not given we may consider the equation

as initially the concentration of OH^- = 0.125M OH^-

pH = 13.09

to find the equivalence point we know that moles of acid = moles of base

moles (n) = C x V

where C = concentration and V is volume in L

n = 0.125M OH^- x 0.0234 L = 0.00292 moles OH^- initially present

So at equilibrium to neutralize 0.00292 moles base we require 0.00292 moles of acid added

so volume of HCl at equilvalence point is

0.00292 moles H⁺ = 0.105M H⁺ x V

V = 0.0278 L = 27.8 mL

so at the equivalence point 27.8 mL of 0.105M HCl was present as we have added 4.1 mL the moles of HCl is

n = C x V

n = 0.105M H⁺ x 0.0041L

n = 0.000430 moles of H⁺ added

As our inital moles of base was 0.00292 moles. By adding 0.000430 moles of acid we neutralize the same moles of base. so the OH remaining can calculated by substracting it with the initial

0.00292 moles - 0.000430 moles = 0.00249 moles OH^- left

so the new concentration will be

0.0234 L + 0.0041 L = 0.0275 L

n = C x V

0.00249 moles = C x 0.0275 L

C = 0.090 M

pH = 12.95

So pH after adding 4.1 mL of 0.105M HCl is 12.95