Question

Please help: A 0.200 M NaOH solution was used to titrate a 18.25 mL HF

Answer 1

The molar concentration of the original HF  solution : 0.342 M

Further explanation

Given

31.2 ml of 0.200 M NaOH

18.2 ml of HF

Required

The molar concentration of HF

Solution

Titration formula

M₁V₁n₁=M₂V₂n₂

n=acid/base valence (amount of H⁺/OH⁻, for NaOH and HF n =1)

Titrant = NaOH(1)

Titrate = HF(2)

Input the value :

$\tt 0.2\times 31.2\times 1=M_2\times 18.25\times 1\\\\M_2=0.342$