Question: Consider the titration of 30.00 ml of .360 M. H2C6H6O6 (abscorbic acid; K1= 6.8e-5; K2=2.8E-12) solution with .280M NaOH. Note the weak acid, H2C6H6O6, is being titrated with the strong base, sodium hydroxide. the neutralization reaction are: H2C6H6O6 + NaOH goes to NaHC6O6 + H20 and NaHC6H6O6 + NaOH goes to Na2C6H6O6 + H2O. A) Compute the pH after 18.0 ml. of NaOH have been added.

Not sure what numbers go where and how to do the problem? Do you need an ICE table?

The first order of business is to determine exactly where you are on the titration curve after 18.0 mL NaOH. A starting point is to determine exactly where the first and second equivalence points are along the titration curve.

How much acid do we have? M x L = mols = 0.360 x 0.030 = 0.0108 moles acid.
So the first eq point will come at 0.0108 moles of NaOH and the second one will come after another 0.0109 moles NaOH.
moles NaOH = 0.0108 = M x L. M = 0.280. Solve for L and I get something like 38.57 mL. Therefore, we know that 18.0 mL is between the starting point and the first eq point. The equation we are working with is something like this.
H2A + NaOH ==> NaHA + H2O
Yes, now use the ICE table to determine moles start, moles NaOH added (at 18.0 mL), moles NaHA formed, moles H2A remaining. Since this is before the first eq point (if I've not done something wrong), you will have formed some of the salt and you will have some of the acid remaining. Aha! a buffer. Use the Henderson-Hasselbalch equation to solve for the pH.

What does HBr(aq) + Mg(HSO3)2 produce?

To determine the pH after 18.0 ml of NaOH has been added, we need to consider the neutralization reactions and the equilibrium equations involved.

First, let's write the balanced chemical equations for the neutralization reactions:

1. H2C6H6O6 + NaOH → NaHC6O6 + H2O
2. NaHC6H6O6 + NaOH → Na2C6H6O6 + H2O

The goal is to find the pH, so we need to consider the concentration of H+ ions, which is related to the acidic species in solution, namely H2C6H6O6 (ascorbic acid) and NaHC6H6O6 (sodium ascorbate).

Now, let's set up an ICE table for the first neutralization reaction:

Reaction: H2C6H6O6 + NaOH → NaHC6O6 + H2O

Initial moles: 0.0360 M * 0.0300 L = 0.00108 mol H2C6H6O6
0.00028 M * 0 L = 0 mol NaOH

Change in moles: -x (as H2C6H6O6 and NaOH react completely, producing x moles of NaHC6O6 and H2O)

Equilibrium moles: 0.00108 mol - x mol H2C6H6O6
x mol NaHC6O6
0 mol NaOH

Using the equilibrium moles, we can calculate the concentrations in terms of x:

[H2C6H6O6] = (0.00108 mol - x) / (0.0180 L) = (0.00108 - x) / 0.018
[NaHC6O6] = x / 0.018

The next step is to consider the equilibrium constant for this reaction, which is given as K1 = 6.8e-5. The equilibrium expression is:

K1 = [NaHC6O6] / [H2C6H6O6]

Now, substitute the concentrations into the equilibrium expression:

6.8e-5 = x / (0.00108 - x)

Now, we can solve for x using the quadratic equation or an approximation method such as the Henderson-Hasselbalch equation. However, in this case, we can make an assumption that the value of x is very small compared to 0.00108 (initial moles of H2C6H6O6). Therefore, we can neglect x in the denominator of the expression:

6.8e-5 = x / 0.00108

Now, solve for x:

x = 6.8e-5 * 0.00108 ≈ 7.34e-8 mol

Since we assumed x to be small, the approximation holds.

Now, we have the moles of NaHC6O6 produced.

Next, we'll determine the number of moles of NaOH that reacted in the second reaction:

2. NaHC6H6O6 + NaOH → Na2C6H6O6 + H2O

The stoichiometry of this reaction indicates that for every one mole of NaHC6H6O6, one mole of NaOH reacts to form one mole of Na2C6H6O6.

Therefore, the moles of NaOH that reacted are also approximately equal to 7.34e-8 mol.

Now, let's calculate the moles of excess NaOH that remain after reacting 18.0 ml:

Initial moles of NaOH = 0.00028 M * 0.018 L = 5.04e-6 mol
Moles of NaOH that reacted = 7.34e-8 mol

Moles of excess NaOH = Initial moles of NaOH - Moles of NaOH that reacted
= 5.04e-6 mol - 7.34e-8 mol
≈ 4.96e-6 mol

Now, let's determine the final volume of the solution:

Initial volume of solution = 30.00 ml
Volume of NaOH added = 18.0 ml

Final volume of the solution = Initial volume of solution + Volume of NaOH added
= 30.00 ml + 18.0 ml
= 48.00 ml = 0.048 L

Finally, we can calculate the concentration of the excess NaOH:

[NaOH] = Moles of excess NaOH / Final volume of solution
= (4.96e-6 mol) / (0.048 L)
≈ 0.1033 M

Since NaOH is a strong base, it dissociates completely, so the concentration of OH- ions is the same as the concentration of NaOH in this case.

Now that we have the concentration of OH- ions, we can calculate the pOH using the formula:

pOH = -log10[OH-]

pOH = -log10(0.1033)
≈ 0.987

Finally, to find the pH, we use the equation:

pH = 14 - pOH

pH = 14 - 0.987
≈ 13.01

Therefore, the pH after adding 18.0 ml of NaOH is approximately 13.01.