Finding pH ofd solution that NaOH is titrated with HCl

Question

(a) Calculate the pH of the solution resulting when 25.0 ml of 0.100 M NaOH is titrated with 0.0, 10.0, 12.5, and 15.0 ml of 0.200 M HCl.

(b) Use the pH values calculated in 6(a) above, and any others you wish to calculate, to plot the curve for the titration on graph paper (pH on y‐axis, volume of HCl on x‐axis). Hint: This graph cannot be done by Excel. You may need to consult a text and/or plot some additional points to get a sense of the sigmoid shape.

(c) Use your plotted titration curve to determine which of the following indicators can be used to indicate the endpoint: Phenolphthalein (8.2 – 9.8); bromothymol blue (6.0 – 7.6); bromocresol green (3.8 – 5.4). Select all that apply.

Answer 1

NaOH is a strong base and HCl is a strong acid. When a strong acid and base are mixed together, they will react 100% to form a neutral salt and water. Thus, the pH of the resultant solution will be determined based on the amount of excess reagent. When doing pH calculations for titrations, the most commonly made mistake is forgetting to take into account the change in volume. As the acid is being added into the base, not only is the OH- being consumed by H+, but the solution is also being diluted. Thus, it will be easier for us to think in terms of moles instead of Molarity and then account for the volume at the end.

A.

When 0.0 mL of 0.200 M HCl is added, then the pH of the solution will simply be the pH of the NaOH solution. NaOH is a strong base so it dissociates 100% to form Na+ and OH-.

pOH = -log[OH-]

pOH = -log(0.100)

pOH = 1.000

pH = 14 - pOH

pH = 14 -1.000 = 13.000 (3 sig figs)

When 10.0 mL oh HCl is added:

The moles of base = 0.0250 L * 0.100 M = 0.00250 moles OH-

The moles of acid = 0.0100 L * 0.200 M = 0.00200 moles H+

When they neutralize, all the acid (limiting reagent) will be consumed leaving 0.00050 moles of OH- in excess.

The total volume = 0.0250 L of NaOH + 0.0100 L of HCl = 0.0350 L of solution.

The concentration of OH- = 0.00050 moles / 0.035 L = 0.0143 M OH-

pOH = -log[OH-] = -log(.0143)

pOH = 1.845

pH = 14 - 1.845 = 12.155

Using this same method, you should be able to calculate the pH at the other two volumes. 12.5 mL of HCl will be the equivalence point, pH = 7 (because it's a strong acid, strong base titration). 15.0 mL of HCl is past the endpoint and should be a pH less than 7 (probably between 2-3).

B. Plot "mL of acid added" on the x-axis and "pH" on the y-axis. The graph should look like an upside-down "S" shape. In other words, the graph will be relatively flat at the beginning, become very steep near the endpoint, and then flatten out again at the bottom. I would recommend doing an additional pH calculation of 25 mL of HCl added to see the second flat region.

C. The choice of indicator depends on what the pH range is during the steep portion of the titration curve. While all 3 indicators could probably be used with reasonable accuracy to mark the endpoint of the titration, bromothymol blue will be the best since it changes colors at the endpoint (in this case, when pH =7). If you have a strong acid - weak base or a weak acid - strong base titration, you'll want to choose a different indicator based on the pH of the endpoint.