Calculate the pH of a solute where 24.5 mL of 0.1 mol/L HCl is added to 25.0 mL of 0.1 mol/L NaOH.

How much excess OH is there? .5ml .1M or
5E-5 moles. That is in a volume of 49.5ml

So if OH concentration is that moles/volume, what is the H+ concentration? Remember the product is 1E-14.

I get a pH of almost 11

To calculate the pH of the solute, we need to determine the excess hydroxide ion (OH-) concentration.

Given that 24.5 mL of 0.1 mol/L HCl is added to 25.0 mL of 0.1 mol/L NaOH, we have a neutralization reaction between the H+ ions from HCl and the OH- ions from NaOH:

HCl + NaOH → NaCl + H2O

Since the reaction is a 1:1 ratio, all of the H+ ions from HCl will react with the OH- ions from NaOH. Therefore, there will be no excess H+ ions.

To find the excess OH- concentration, we need to determine the total number of moles of OH- provided by NaOH. The moles of OH- can be calculated as follows:

(0.1 mol/L) × (0.025 L) = 0.0025 mol

This means that there are 0.0025 mol of OH- ions in the 25.0 mL of NaOH solution.

Since the total volume of the solution after mixing is 49.5 mL (24.5 mL + 25.0 mL), we can calculate the concentration of OH- ions as follows:

0.0025 mol / 0.0495 L = 0.0505 mol/L = 5.05 × 10^-2 mol/L

Now, we can use the relationship between H+ and OH- concentrations in water, which is given by the equation:

[H+] × [OH-] = 1.0 × 10^-14

Since we know the OH- concentration is 5.05 × 10^-2 mol/L, we can rearrange the equation to solve for the H+ concentration:

[H+] = 1.0 × 10^-14 / 5.05 × 10^-2 = 1.98 × 10^-13 mol/L

Finally, we can calculate the pH using the equation:

pH = -log[H+]

pH = -log(1.98 × 10^-13) ≈ 12

Therefore, the pH of the solute in this solution is approximately 12, not almost 11 as previously mentioned.