Sodium carbonate (0.3379 g, FW: 105.99) was dissolved in deionized water to give a solution with a total volume of 250.0 mL. What was the pH of the resulting solution

pH = 11.179

Explanation:

The sodium carbonate, Na2CO3, is in equilibrium with water as follows:

Na2CO3 + H2O ⇆ NaHCO3 + OH⁻

Where Kb is 1.786x10⁻⁴:

Kb = 1.786x10⁻⁴ = [NaHCO3] [OH-] / [Na2CO3]

Where [NaHCO3] = [OH-] = X, our incognite

And [Na2CO3] is molar concentration of sodium carbonate:

[Na2CO3] = 0.3379g * (1mol / 105.99g) = 3.188x10⁻³ moles Na2CO3

In 250.0mL = 0.2500L:

3.188x10⁻³ moles Na2CO3 / 0.2500L = 0.01275M

Replacing in Kb expression:

1.786x10⁻⁴ = [NaHCO3] [OH-] / [Na2CO3]

1.786x10⁻⁴ = [X] [X] / [0.01275]

2.278x10⁻⁶ = X²

X = [OH⁻] = 1.509x10⁻³M

As pOH = -log [OH⁻] = 2.821

And pH = 14 - pOH

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.

the answer would be c