What is [OH^-] for a solution at 25°C that has [H₃O⁺] = 8.23 × 10^-2^M? A) 8.23 × 10^-2^M B) 1.22 × 10^-6^M C) 8.23 × 10^-12^M D) 1.22 × 10^-13^M E) 8.23 × 10^-16^M

Question

Quizplus · Accepted Answer

First, we need to identify the species being referred to in the given notation:
- HS1U1B13S1U1B0OS1U1P1+S1S1P0 is the notation for bisulfite ion (HSO3-) which has a charge of -1.
- S1S1P0 is the notation for hydrogen ions (H+) which have a charge of +1.

Using this information, we can rewrite the given notation as:
[OHSO3-][H+]^8.23 × 10^-12

Using the rules of the law of mass action, we can write the equilibrium expression for the reaction involving these species:

HSO3- + H2O ⇌ SO32- + H3O+

The expression for this reaction is:

K = [SO32-][H3O+] / [HSO3-]

Since we are given the concentration of bisulfite ion and hydrogen ions, we can substitute these values into the equilibrium expression:

K = ([H+][SO32-]) / [HSO3-] = [H+]^1[SO32-]^1 / [HSO3-]^1

We can see that the stoichiometric coefficients of all species in the reaction are 1. Therefore, the value of K is equal to the value of the equilibrium expression, which we can equate to the given [OHS1U1P1-S1S1P0]:

K = [H+]^1[SO32-]^1 / [HSO3-]^1 = 8.23 × 10^-12

Taking the negative logarithm of both sides, we get:

-pH + pSO32- = 11.08

We are asked to solve for the pH of the solution. To do this, we need to know the concentration of sulfite ion ([SO32-]). We can find this by rearranging the equilibrium expression and substituting in the given values:

[SO32-] = K[HSO3-] / [H+]

[SO32-] = (8.23 × 10^-12)(1) / 8.23 × 10^-12
[SO32-] = 1 M

Substituting this value into our equation, we get:

-pH + p(1) = 11.08

-pH + 0 = 11.08

pH = -11.08

However, pH is not negative. This is because we made an assumption earlier that [H+] = [SO32-]. This is not necessarily true. In fact, in most cases, [H+] is much smaller than [SO32-]. Therefore, we need to make an approximation:

[HSO3-] ≈ [SO32-]
[H+] << [SO32-]

Using these approximations, we can simplify the equilibrium expression:

K ≈ [SO32-]^2 / [HSO3-]

[SO32-]^2 / [HSO3-] = 8.23 × 10^-12

[SO32-] = sqrt(8.23 × 10^-12[HSO3-])

[SO32-] = sqrt(8.23 × 10^-12(1))

[SO32-] = 9.08 × 10^-6 M

Since [H+] << [SO32-], we can assume that [H+] ≈ 0. Therefore, the pH of the solution is approximately equal to the pKa of the bisulfite ion, which is 7.2. Using the relationship between pH and pKa, we can calculate the hydrogen ion concentration:

pH = pKa + log([base]/[acid])

7.2 = log([SO32-]/[HSO3-])

[SO32-]/[HSO3-] = 10^7.2

[SO32-]/[HSO3-] = 6.31 × 10^7

[SO32-] = (6.31 × 10^7)[HSO3-]

[SO32-] = (6.31 × 10^7)(1)

[SO32-] = 6.31 × 10^7 M

Substituting this value into our equation, we get:

-pH + p(6.31 × 10^7) = 11.08

-pH + 7.8 = 11.08

pH = 3.28

Therefore, the best choice is D) 1.22 × 10^-13 S1S1P0S1U1P1MS1S1P0.

What Is [OH-] for a Solution at 25°C That Has

What Is [OH^-] for a Solution at 25°C That Has