We will use this two reaction equation:
H2SO3 + H2O ↔ H3O+ + HSO3- Ka1 = 1.3 x 10^-2
HSO3- + H2O ↔ H3O+ + SO3 2- Ka2= 6.3 x 10^-8
we will use the ICE table for the first equation:
H2SO3 + H2O ↔ H3O+ + HSO3-
initial 0.025 0 0
change -X +X +X
Equ (0.025-X) X X
Ka1 = [H3O+] [HSO3-] / [H2SO3]
1.3 x 10^-2 = X^2 / (0.025 - X) by solving for X
∴ X = 0.0127
when [H3O+] = X
∴[H3O+] = 0.0127 M
and when [HSO3-] = X
∴[HSO3-] = 0.0127 M
and when [H2SO3] = 0.025 - X
∴[H2SO3] = 0.025 - 0.0127
= 0.0123 M
when Kw = [OH-][H3O+]
and Kw = 1.1 x 10^-14 / 0.0127
∴[OH-] = 1.1 x 10^-14 / 0.0127
= 8.66 x 10^-13 M
- by using the ICE table for the second equation:
HSO3- + H2O ↔ H3O+ + SO3 2-
initial 0.0127 0.0127 0
change -X +X +X
Equ (0.0127-X) (0.0127+X) X
when Ka2 = [SO32-] [H3O+] / [HSO3-]
by substitution:
6.3 x 10^-8 = X(0.0127+X) / (0.0127-X)
as the Ka2 is so small so we can assume that (0.01271 + X) & (0.01271-X) = 0.01271 and neglect X
6.3 x 10^-8 = 0.0127X /0.0127
∴X = 6.3 x 10^-8
when [SO3 2-] = X
∴[SO32-] = 6.3 x 10^-8
Answer:
False
Explanation:
Increase in polarity of a molecule leads to higher boiling points. The more polar a molecule is, the higher the energy required to breaks intermolecular forces of attraction hence the higher the boiling point. This is the reason why ionic compounds and compounds having polar covalent bonds in them tend to have high boiling points.
Answer:
habitat destruction I think
Explanation:
Answer:
The mass of NaCl is 0.029 grams
Explanation:
Step 1: Data given
Molecular weight of NaCl = 58.44 g/mol
Volume of solution = 100 mL = 0.100 L
Molarity = 0.0050 M
Step 2: Calculate moles NaCl
Moles NaCl = molarity * volume
Moles NaCl = 0.0050 M * 0.100 L
Moles NaCl = 0.00050 moles
Step 3: Calculate mass NaCl
Mass NaCl = moles NaCl * molar mass NaCl
Mass NaCl = 0.00050 moles * 58.44 g/mol
Mass NaCl = 0.029 grams
The mass of NaCl is 0.029 grams
