What is the concentration of H₃O⁺ and ^-OH in pure water at room temperature? A) 1.0 *10^-¹⁴ M B) 1.0 *10^-⁷ M C) 1.0 M D) 1.0 * 10⁷ M E) 1.0 *10¹⁴ M

Question

Quizplus · Accepted Answer

The question is asking for the concentration of two species in pure water at room temperature. We can use the ionization constants (Kas) of the given species to calculate their concentrations.
HS1U1B13S1U1B0OS1U1P1+S1S1P0 is an acidic species, so we can use Ka to calculate its concentration. The balanced equation for its ionization is:
HS1U1B13S1U1B0OS1U1P1+S1S1P0 ⇌ H+ + S1U1B13S1U1B0OS1U1P1- + S1S1P0
Ka = [H+][S1U1B13S1U1B0OS1U1P1-]/[HS1U1B13S1U1B0OS1U1P1+S1S1P0]
At equilibrium, the concentration of HS1U1B13S1U1B0OS1U1P1+S1S1P0 will be equal to its initial concentration (since we are dealing with pure water). Therefore, we can assume that [HS1U1B13S1U1B0OS1U1P1+S1S1P0] = x, [H+] = x, and [S1U1B13S1U1B0OS1U1P1-] = [S1S1P0] = x.
Substitute these values into the Ka expression and solve for x:
Ka = x*x/x = x
Ka = 1.0 x 10^-7 (from a table of ionization constants)
x = 1.0 x 10^-7 M
So, the concentration of HS1U1B13S1U1B0OS1U1P1+S1S1P0 is 1.0 x 10^-7 M in pure water at room temperature.
Similarly, S1U1P1-S1S1P0OH is a basic species, so we can use Kb to calculate its concentration. The balanced equation for its ionization is:
S1U1P1-S1S1P0OH + H2O ⇌ S1U1P1+ + OH- + H2O
Kb = [S1U1P1+][OH-]/[S1U1P1-S1S1P0OH]
At equilibrium, the concentration of S1U1P1-S1S1P0OH will be equal to its initial concentration (since we are dealing with pure water). Therefore, we can assume that [S1U1P1-S1S1P0OH] = x, [S1U1P1+] = x, and [OH-] = x.
Substitute these values into the Kb expression and solve for x:
Kb = x*x/x = x
Kb = 1.0 x 10^-7 (from the same table of ionization constants)
x = 1.0 x 10^-7 M
So, the concentration of S1U1P1-S1S1P0OH is also 1.0 x 10^-7 M in pure water at room temperature.
The only answer choice that includes both of these concentrations and is correct is B.

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What Is the Concentration of H₃O⁺ and ^-OH in Pure