Can you help me with the following: Calculate the pH that results when 25 mL of .1M HCL is titrated with .1M NaOH solution run in from a buret at each of the stages of volume of .1 M NaOH add in 5.0 mL incrememnts beginning with 0 mL of NaOH

HCl + NaOH ==> NaCl + H2O

At zero mL. YOu have 0.1 M HCl so the pH is pH = -log(H^+). Since HCl is a strong acid, H^+ = 0.1 M and pH = 1.

All of the others are done this way:
a. moles HCl initially = M x L = 0.025 x 0.1 = 0.0025 moles HCl.
b. moles NaOH added. (at 5 mL this is 0.005L x 0.1 M = 0.0005 moles NaOH.)
c. Moles HCl are neutralized by moles NaOH. Start with 0.0025 HCl - 0.0005 NaOH = 0.0020 moles HCl left. (H^+) = 0.0020moles/L (25 mL from HCl + 5 mL of added NaOH = 30 mL = 0.030 L)
d.pH = -log(H^+). Substitute remaining H^+ and solve.

All of the points work the same way until reaching the equivalence point which is at 25 mL NaOH. Here the pH is determine by the hydrolysis of the salt. Neither Na^+ nor Cl^- are hydrolyzed; therefore, the pH is 7 from pure water.

Points after the equivalence point are just excess NaOH. Therefore, calculate excess OH and pOH from that and convert to pH.

what is the pH of a solution obtained by mixing 70 mL of 0.125M NaOH with 30ML of 0.105M of HCL

Sure! To calculate the pH at each stage of the titration, we need to consider the reaction between HCl (a strong acid) and NaOH (a strong base). The reaction is as follows:

HCl + NaOH -> NaCl + H2O

At each stage, we will determine the moles of HCl and NaOH, calculate the excess or remaining moles of either of the reactants, and then calculate the pH using the concentration of the remaining reactant.

Let's go step by step:

1. Initial stage: 0 mL of NaOH added
- Moles of HCl = (0.025 L) x (0.1 mol/L) = 0.0025 mol (since the initial volume of HCl is 25 mL)
- Moles of NaOH = 0 (since none has been added yet)
- Excess HCl: 0.0025 mol
- Calculate the pH using the excess HCl.

2. After adding 5.0 mL of NaOH:
- Moles of HCl = (0.025 L) x (0.1 mol/L) = 0.0025 mol
- Moles of NaOH = (0.005 L) x (0.1 mol/L) = 0.0005 mol
- Remaining HCl: 0.0025 - 0.0005 = 0.0020 mol
- Calculate the pH using the remaining HCl.

3. Repeat the above steps for each subsequent 5.0 mL increment of NaOH until the titration is complete.

Keep in mind that during the titration, the base is neutralizing the acid. The pH will depend on the ratio of remaining acid to remaining base. As the reaction progresses, the ratio changes, which impacts the pH.

To calculate the pH, you can use the formula:

pH = -log10([H+])

In this case, [H+] represents the concentration of the remaining acid (HCl) after each stage of the titration.

I hope this helps! Let me know if you have any further questions.