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One proposed solution to ocean acidification is to, essentially, add a base back to the ocean. If we were to try to change the pH of the Atlantic Ocean (roughly 3 x 1020 L) to 8.2, how many grams of NaOH would we have to add if there is a 2.3 mM carbonic acid buffer current at a pH of 8.1?

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Answer:

0.0002 g

Step-by-step explanation:

A buffer is a solution in which is presented an equilibrium between a weak acid and its conjugate base (or between a weak base and its conjugate acid). Because of the equilibrium, when an acid or base is added to the solution, the pH remains almost unaltered (if an acid is added, the base reacts with it, and if a base is added the acid reacts with it).

However, at some point, the buffer will not function, because there'll be no more acid or base to react. So, if the acid is HA, and the conjugate base A⁻, the equilibrium reaction is:

HA ⇄ H⁺ + A⁻

The pH of the solution can be determined by the Handerson-Halsebach equation:

pH = pKa + log[A⁻]/[HA]

Where pKa = -logKa, and Ka is the equilibrium constant of the acid. This acid is a polyprotic acid, which means that it has 2 hydrogens to lose, and so it has two equilibriums, with pKa values, pKa1 = 6.36 and pKa2 = 10.25. Because the pH of the solution (8.2) is close to pKa1, then the equilibrium that predominates is the first one.

8.1 = 6.36 + log[A⁻]/[HA]

log[A⁻]/[HA] = 1.74

[A⁻]/[HA] =

[A⁻]/[HA] = 54.954

Because the volume is the same, we can use the number of moles instead of the molarity. After dissociation, the number of moles of both substances must be equal to the initial one ( 2.3 mM) so:

A⁻ + HA = 2.3

A⁻ = 2.3 - HA

(2.3 - HA)/HA = 59.954

59.954HA = 2.3 - HA

60.954HA = 2.3

HA = 0.038 mM

So, when NaOH is added to it, it must consume H⁺, and so more HA will dissociate, and more [A⁻] will be formed. So, with the new pH:

8.2 = 6.36 + log[A⁻]/[HA]

log[A⁻]/[HA] = 1.84

[A⁻]/[HA] =

[A⁻]/[HA] = 69.183

Because the OH- of NaOH reacts with H+, A⁻ + HA will not change, so:

(2.3 - HA)/HA = 69.183

69.183HA = 2.3 - HA

70.183HA = 2.3

HA = 0.033 mM

The reaction between OH- and H+ is a 1:1 reaction, and so, the dissociation reaction is a 1:1:1 reaction. Thus, the number of moles of HA consumed is the number of moles of H+ that was consumed, which are the number of moles of NaOH added:

n = 0.038 - 0.033 = 0.005 mM = 5.0x10⁻⁶ mol

NaOH has a molar mass equal to 40 g/mol, and the mass is the number of moles multiplied by it, so:

m = 5.0x10⁻⁶ * 40 = 0.0002 g

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