Consider the titration of 50 mL of 0.250 M HCl with 0.1250 M NaOH. Calculate the pH of the resulting solution after the following volumes of Na OH have been added.

a) 0.00 mL
b) 50.00 mL
c) 99.90 mL
d) 100.00 mL
e) 100.1 mL

The question is, I get, for example, 0.9 pH for "a." However, where I am getting the molarity to do this is from the NaOH solution. Am I right? If not, from which solution should I get the molarity from?

For a. You have 0.00 mL of NaOH; therefore the pH will be determined solely by HCl.

pH = -log(0.250)
pH = 0.602

To calculate the pH of the resulting solution during a titration, you need to consider the reaction between the acid (HCl) and the base (NaOH). The balanced chemical equation is:

HCl + NaOH → NaCl + H2O

In this reaction, the acid and base react in a 1:1 mole ratio, which means that for every mole of HCl, you need 1 mole of NaOH to completely neutralize it.

Given that the initial volume of HCl is 50 mL and its molarity is 0.250 M, you can calculate the number of moles of HCl using the formula:

moles = volume (L) × molarity (mol/L)

So, the number of moles of HCl initially is:
moles HCl = (50 mL) × (0.250 mol/L) = 0.0125 mol

Now, let's go through each situation and determine the pH of the resulting solution after adding different volumes of NaOH:

a) 0.00 mL NaOH:
At this point, no NaOH has been added, and the volume of HCl remains unchanged. Hence, the number of moles of HCl is still 0.0125 mol. To calculate the pH, you need to know the volume of the final solution. Assuming the total volume remains constant at 50 mL, you can use the Henderson-Hasselbalch equation to calculate the pH:

pH = -log10[H+]
[H+] = moles HCl / volume (L)

From the calculation above, the number of moles of HCl is 0.0125 mol, and assuming a constant volume of 50 mL, the pH can be calculated as follows:

pH = -log10(0.0125 mol / 0.050 L) = -log10(0.25) = 0.60

b) 50.00 mL NaOH:
At this point, you have added an equal volume (50 mL) of NaOH to HCl. The balanced chemical equation shows that the reaction has reached a 1:1 mole ratio between HCl and NaOH, meaning that all the HCl has been neutralized. Therefore, the moles of unreacted HCl are now 0 mol. Since there are no H+ ions left in the solution, the resulting solution is neutral, and the pH is 7.

c) 99.90 mL NaOH:
At this point, you have almost reached the equivalence point, where the moles of NaOH added are nearly equal to the moles of HCl initially present. The pH can be calculated in the same way as in (a), using the moles of HCl and the total volume of the solution after adding 99.90 mL of NaOH.

d) 100.00 mL NaOH:
At this point, the moles of NaOH added are exactly equal to the moles of HCl initially present. You have reached the equivalence point. As the reaction between HCl and NaOH is complete, there are no H+ ions or OH- ions remaining in the solution. Therefore, the resulting solution is neutral, and the pH is 7.

e) 100.1 mL NaOH:
At this point, you have added slightly more NaOH than is required to neutralize all the HCl. The excess NaOH will contribute OH- ions to the solution, making it basic. To calculate the pH, you need to determine the number of moles of NaOH added in excess and consider the additional OH- ions present.

To determine the moles of NaOH added in excess, you can subtract the moles of HCl from the moles of NaOH:

moles NaOH in excess = moles NaOH added - moles HCl

Then, you can calculate the concentration of OH- ions in the solution using the moles of NaOH in excess and the total volume of the solution after adding 100.10 mL of NaOH. Finally, use this concentration value to calculate the pOH and then convert it to pH using the equation:

pH = 14 - pOH

Keep in mind that you need to consider the volume of the resulting solution after adding NaOH to calculate the pH accurately.

In summary, the molarity to calculate the pH in a titration should be obtained from the solution of the acid being titrated, HCl in this case. The concentration of the base (NaOH) is used to determine the number of moles of the base added at each point in the titration, which is then used to calculate the pH based on the reaction stoichiometry and volume.