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

Question

Quizplus · Accepted Answer

First, we need to break down the long chemical formula into its individual components:
HS1U1B13S1U1B0O = bisulfite ion
S1U1P1+ = hydrogen sulfite ion
S1S1P0 = sulfite ion
S1U1P1- = hydrogen sulfate ion
+ = positively charged sodium ion
When we add these up, we get the overall equation for the reaction in 0.10 M NaOH:
HSO3- + HSO3- + SO32- + HSO4- + Na+ + OH- → H2O + Na+ + SO42-
Notice that the sodium ions and hydroxide ions simply act as spectator ions and do not affect the concentration of the other ions. 
Now, we need to figure out the initial concentrations of each of the ions. Since we are starting with 0.10 M NaOH, we know that the initial concentration of OH- is also 0.10 M. The other ions all come from the bisulfite ion, which we can assume fully dissociates in solution. Therefore, the initial concentrations of each of the ions are:
[HSO3-] = 2 × 0.10 M = 0.20 M
[SO32-] = 1 × 0.10 M = 0.10 M
[HSO4-] = 1 × 0.10 M = 0.10 M
Now, we can set up an ICE table to determine the equilibrium concentrations. Since we know that the overall reaction consumes two hydrogen sulfite ions and one sulfite ion for every bisulfite ion, we can let x represent the change in molarity for each of these ions. For the hydrogen sulfate ion and the sodium ion, the change in molarity is equal to the concentration of hydroxide ions added, which is also equal to x. The ICE table looks like this:
HSO3- + HSO3- + SO32- + HSO4- + Na+ + OH- → H2O + Na+ + SO42-
I:  0.20 M      0.20 M      0.10 M      0.10 M     0.10 M    0.10 M
C: -2x          -2x           -x           -x          +x        +x
E:  0.20 - 2x 0.20 - 2x  0.10 - x  0.10 - x  0.10 + x  0.10 + x
At equilibrium, the overall concentration of hydrogen sulfite ion is the sum of the concentrations of each individual hydrogen sulfite ion, so we add 0.20 - 2x + 0.20 - 2x = 0.40 - 4x. Similarly, the overall concentration of hydrogen sulfate ion is 0.10 + x, since there is only one hydrogen sulfate ion produced for each hydroxide ion. The overall concentration of sulfite ion is 0.10 - x. Finally, the concentration of sodium ion and hydroxide ion at equilibrium is simply the initial concentration plus the change in molarity from each ion. 
The equation for the equilibrium reaction is:
HSO3- + HSO3- + SO32- + HSO4- + Na+ + OH- → H2O + Na+ + SO42-
0.40 - 4x × 0.10 + x × 0.10 - x × 0.10 + x × 0.10 + x × 0.10 = Kw/[OH-]
where Kw is the ion product constant for water, which equals 1.0 × 10-14 at 25°C. Solving for x, we get x = 1.3 × 10-13. Therefore, the equilibrium concentrations are:
[HSO3-] = 0.20 - 2x = 0.20 M (to three significant figures)
[SO32-] = 0.10 - x = 0.10 M (to three significant figures)
[HSO4-] = 0.10 + x = 0.10 M (to three significant figures)
Therefore, the concentration of HS1U1B13S1U1B0OS1U1P1+S1S1P0 at equilibrium is:
[HSO3-] + [HSO4-] + [SO32-] + [H2SO3] 
= 0.20 M + 0.10 M + 0.10 M 
= 0.40 M. 
Converting to scientific notation, the concentration is 4.0 × 10S1U1P1-1S1S1P0 M. The answer is therefore E.

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