Question

A 20.00 mL solution of HClO4 of unknown concentration is titrates with 33.55mL of 0.09945M NaOH solution.

A) calculate the concentration of the unknown HClO4 solution

B) whay is the pH of the solution before titration was started?

C) what is the pH of the solution when 75.00% of the acid has been neutralized?

Answer 1

The balanced chemical reaction can be written as:

NaOH + HClO₄ = NaClO₄ + H₂O

1 mole NaOH reacts with 1 mole HClO₄ completly

moles of NaOH = molarity x volume in litres

= 0.09945 M x 0.03355 L

= 0.0033365475 moles

so if HClO4 is completely neutralized it needs exact same amount as NaOH can be seen from reaction equation

0.0033365475 moles = 0.02 L x M₁(molarity of HClO4)

M₁ = 0.166827375 M

(A) = 0.169 M

before titration

pH = -log[H+] = -log[0.169]

(B) pH = 0.772

for 75% neutralization

0.02 x 0.169M =0.0338 moles of HClO4 total

75% of this is = 0.0338 x0.75 = 0.02535 moles

total volume = volume of NaOH + HClO4 = 20 mL + 33.55 mL = 53.55 mL

molarity = 0.02535 moles/0.05355 L = 0.4734 M

pH = -log[H+] = 0.32

comment back if you need further help

please revert if you need further help