A0.10 m solution of a weak monoprotic acid has a ph of 3.40 at 25°c. what is the acid-ionization
constant, ka, for this acid?
a) 1.6 x 10-6
b) 4.0 x 10-4
c) 3.4 x 10-5
d) 1.2 x 10-3
e) 1.8 x 10-7

The correct answer is A) 1.6 x 10-6

Explanation:

A weak monoprotic acid has the following dissociation equilibrium. At the beggining (t=0), the concentration of the monoprotic acid (HA) is equal to 0.10 M and the concentration of the ions H⁺ and A⁻ is zero (no dissociation). At a time t, dissociation occur and there is x concentration of H⁺ and A⁻ which is given by the dissociation constant Ka.

           HA(aq)    ⇄     H⁺(aq)    +       A⁻(aq)

t=0        0.10 M              0                     0

t               -x                    +x                  +x

eq          0.10 M-x             x                   x

Ka=

As the pH is 3.40, we can calculate the concentration of both H⁺ and A⁻, as follows:

pH= - log (conc H⁺)= -log x

⇒ x = = 3.98 x 10⁻⁴

Now we introduce x in the previous equation to calculate Ka:

Ka=

Ka= 1.59 x 10⁻⁶ ≅ 1.60 x 10⁻⁶

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.