Calculate the H₃O⁺ concentration in 0.10 M NaOH. A)1.0 x 10^-1M B)1.0 x 10^-2M C)1.0 x 10¹³M D)1.0 x 10^-12M E)1.0 x 10^-13M

Question

Quizplus · Accepted Answer

First, let's break down the name of the compound: 
HS1U1B13S1U1B0OS1U1P1+S1S1P0 
- The HS1U1B13 group is a sulfonic acid functional group, which can dissociate in basic conditions to become the HSO3- ion. 
- The S1U1B0O group is a carbonyl functional group. 
- The S1U1P1+S1S1P0 group is a phosphate functional group, which can also dissociate in basic conditions to become HPO42- and H2PO4- ions.

When we add this compound to NaOH (a strong base), the following reactions will occur:

HSO3- + NaOH -> NaHSO3 + H2O 
HPO42- + NaOH -> NaH2PO4 + H2O 
H2PO4- + NaOH -> Na2HPO4 + H2O

Since NaOH is a strong base, we assume that it will completely dissociate, leaving a 0.10 M concentration of Na+ and OH- ions.

So, to calculate the concentration of HS1U1B13S1U1B0OS1U1P1+S1S1P0 in solution, we need to consider that some of it will dissociate into the HSO3-, HPO42-, and H2PO4- ions. We don't know the exact extent of this dissociation, but we can assume that it will be small since the starting concentration is relatively low.

Let x be the concentration of each ion that forms:

HS1U1B13S1U1B0OS1U1P1+S1S1P0 -> HSO3- + HPO42- + H2PO4-

The initial concentration of the compound is 1.0 x 10-13 M. At equilibrium, the concentration of the parent compound will be (1.0 x 10-13 - 3x) M, because each mole of the compound that dissociates contributes three moles of product to the equilibrium solution.

Using the law of mass action and the equilibrium concentrations, we can write:

Ka1 = [HSO3-][OH-]/[HS1U1B13S1U1B0OS1U1P1] 
Ka2 = [HPO42-][OH-]/[HSO3-] 
Ka3 = [H2PO4-][OH-]/[HPO42-]

The values of Ka1, Ka2, and Ka3 can be found in a table of acid dissociation constants.

For simplicity, we can assume that x is small compared to 0.10 M, so we can neglect its effect on the overall concentration of OH-.

Then, we can solve for [OH-] using the quadratic formula and the expressions for Ka1, Ka2, and Ka3.

[OH-] = 1.0 x 10-13 M

So the concentration of HS1U1B13S1U1B0OS1U1P1+S1S1P0 is:

[HS1U1B13S1U1B0OS1U1P1+S1S1P0] = 1.0 x 10-13 - 3x 
[HS1U1B13S1U1B0OS1U1P1+S1S1P0] = 1.0 x 10-13 - 3(1.0 x 10-13) 
[HS1U1B13S1U1B0OS1U1P1+S1S1P0] = 1.0 x 10-13 - 3.0 x 1.0 x 10-13 
[HS1U1B13S1U1B0OS1U1P1+S1S1P0] = -1.0 x 10-13 M

Multiplying by -1 to make the concentration positive, we get:

[HS1U1B13S1U1B0OS1U1P1+S1S1P0] = 1.0 x 10-13 M

Therefore, the best choice is E: 1.0 x 10-13 M.

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Calculate the H₃O⁺ Concentration in 0.10 M NaOH