What is the concentration of hydroxide ions in pure water at 30.0 ∘c, of kw at this temperature is 1.47x10-14?

[OH⁻] = 1.21×10⁻⁷

Explanation:

Water dissociates in the solution as:

H₂O ⇔ H⁺ + OH⁻

For pure water, the concentration of [H⁺] and [OH⁻] are equal.

Kw is defined as the product of the active concentration of  H⁺ and OH⁻ furnished by the water.

So,

Kw = [H⁺] [OH⁻]

Given that, at 30.0 °C, the Kw is 1.47×10⁻¹⁴

So,

Kw = [H⁺] [OH⁻]

As stated above, [H⁺] = [OH⁻]

1.47×10⁻¹⁴ = [OH⁻]²

[OH⁻] = 1.21×10⁻⁷

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.