1) What is the pH of 1.32E-4 M KOH(aq) at 25 °C? ( K w = 1.01E-14) 2)what is the
ID: 948785 • Letter: 1
Question
1) What is the pH of 1.32E-4 M KOH(aq) at 25 °C? (Kw = 1.01E-14)
2)what is the relationship between polarizability and intermolecular forces?
a)Polarizability effects the strength of London Dispersion Forces
b)Polarizability is only relevant in molecules that exist in the liquid phase
c)The polarizability increases as you move to the left of the periodic table
d)Polarizability effects the strength of Dipole-Dipole interactions
e)Polarizability effects the molecular weight of the molecule
a)Polarizability effects the strength of London Dispersion Forces
b)Polarizability is only relevant in molecules that exist in the liquid phase
c)The polarizability increases as you move to the left of the periodic table
d)Polarizability effects the strength of Dipole-Dipole interactions
e)Polarizability effects the molecular weight of the molecule
Explanation / Answer
1) Given,
[KOH] = 1.32 x 10^-4 M
We know that KOH is a strong base and it dissociates completely in aqueous solution as follows,
KOH -----> K+ + OH-
=> [OH-] from KOH = 1.32 x 10^-4 M
Kw = [H+] [OH-] = 1.01 x 10^-14
=> [H+] = 1.01 x 10^-14 / 1.32 x 10^-4 = 7.65 x 10^-11 M
pH = - log [H+] = - log (7.65 x 10^-11) = 10.12
2)
a) Polarizability effects the strength of London Dispersion Forces.
As polarizability increases, the dispersion forces also become stronger.
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