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 :
Determining on the temperature, ice could melt, water could freeze or evaporate. Just an example.
NaCl is made of Na and Cl
The molar mass of Na = 22.989 g/mol
The molar mass of Cl = 35.453 g/mol
Molar mass of NaCl = 22.989 + 35.453
= 58.44 g/mol
According to the reaction equation:
CH3COO- + H+ → CH3COOH
initial 0.25 0.15
change - 0.025 + 0.025
Equ (0.25-0.025) (0.15 + 0.025)
first, we have to get moles acetate and moles acetic acid:
moles of acetate = 0.25 - 0.025 = 0.225 moles
∴ [CH3COO-] = 0.225 mol / 1 L = 0.225 M
moles of acetic acid = 0.15 + 0.025 = 0.175 moles
∴ [ CH3COOH] = 0.175 mol / 1L = 0.175 M
Pka = -㏒ Ka
= -㏒ 1.8 x 10^-5
= 4.74
from H-H equation we can get the PH value:
PH = Pka + ㏒ [acetate / acetic acid]
PH = 4.74 + ㏒[0.225/0.175]
∴ PH = 4.8
