Determine the pH of a solution after 20.00 mL of 0.4963 M HI has been titrated with 12.64 mL of 0.5174 M NaOH.

millimols HI = mL x M = about 9.9 but you need a better answer on this AND all that follows. My numbers are just estimates.

mmols NaOH = approx 6.5

......HI + NaOH ==> NaI + H2O
I....9.9...0.........0.....0
add........6.5...............
C...-6.5..-6.5......6.5...6.5
E... 3.4...0........6.5...6.5

So the solution at the end is 3.4 mmols HI and 6.5 mmols NaI. HI is a strong acid and NaI is neutral in water; therefore, the pH is determined by the HI.
M HI = mmols/mL = ?
Convert that to pH.
Remember to do this more accurately from the beginning. Watch the significant figures.

To determine the pH of a solution after a titration, we need to know the reaction that is occurring between the acid (HI) and the base (NaOH). In this case, HI is a strong acid and NaOH is a strong base, so they will react in a 1:1 mole ratio to produce water (H2O) and a salt (NaI).

First, let's calculate the number of moles of HI and NaOH used in the titration:

Moles of HI = concentration of HI × volume of HI
= 0.4963 M × 0.02000 L
= 0.00993 mol

Moles of NaOH = concentration of NaOH × volume of NaOH
= 0.5174 M × 0.01264 L
= 0.00654 mol

Since HI and NaOH react in a 1:1 mole ratio, the moles of HI and NaOH used are equal. Therefore, the limiting reactant is HI, and all of it will react with NaOH.

Next, let's determine the excess moles of NaOH:

Excess moles of NaOH = moles of NaOH - moles of HI
= 0.00654 mol - 0.00993 mol
= -0.00339 mol

Since the moles of HI and NaOH react completely, the excess moles of NaOH are negative because there is an excess of HI remaining. This means the solution is acidic.

To find the concentration of the excess HI, we need to calculate the remaining volume of the solution after the titration. The total volume can be determined by adding the volumes of HI and NaOH used:

Total volume of solution = volume of HI + volume of NaOH
= 0.02000 L + 0.01264 L
= 0.03264 L

Now, we can determine the concentration of the excess HI:

Concentration of HI = moles of HI / total volume of solution
= 0.00993 mol / 0.03264 L
= 0.304 M

Finally, to determine the pH of the solution, we can use the formula:

pH = -log[H+]

The concentration of H+ (hydronium ions) is the same as the concentration of the excess HI. Thus, the pH can be calculated as:

pH = -log(0.304)
≈ 0.517

Therefore, the pH of the solution after the titration is approximately 0.517.