A 50.0 mL sample of 0.50 M HBr is titrated with 0.25 M KOH. What is the pH after the addition of 60.0 mL of KOH?

A) 4.1
B) 13.4
C) 1.7
D) 2.0
E) 7.7

I'm confused on the process of solving this problem. My work so far:
.050L X .50 mol/1L= .025 mol HBR
.060L X .25 mol/1L= .015 mol KOH

I looked at a similar problem and it said then if HBR was in excess to take mol HBR-mol KOH/total L then the -log of that answer. The answer Im getting doesn't match the multiple choice. Where am I going wrong? Thanks.

mols HBr = 0.025 from your work

mols KOH = 0.015 from yur work.

.......HBr + KOH ==> KBr + H2O
I....0.025..0.015.....0.....0
C...-0.015.-0.015....0.015..0.015
E....0.010....0......0.015
HBr is in excess.
(HBr)= mols/L = 0.010/(0.06+0.05) = approx 0.09
Then pH = -log(H^) = ?
If you're still stuck post your work and I'll find the error.
By the way, when you first posted the problem you could have shown your work and I could have spotted the problem instantly. My best guess is that you didn't convert mols to M. Also if you need more help post the choices available. Sometimes they don't give exact numbers on choices.

To solve this problem, we need to determine the moles of HBr and KOH that react in the titration, and then calculate the concentration of the remaining species to find the pH.

First, convert the volume of the HBr solution to liters:
50.0 mL = 50.0 mL x (1 L / 1000 mL) = 0.050 L

Next, calculate the moles of HBr present in the solution using the concentration of HBr:
moles of HBr = concentration x volume
moles of HBr = 0.50 M x 0.050 L = 0.025 mol

Similarly, calculate the moles of KOH present in the solution using the concentration of KOH and the volume added:
moles of KOH = concentration x volume
moles of KOH = 0.25 M x 0.060 L = 0.015 mol

Since HBr and KOH react in a 1:1 ratio, the moles of HBr remaining after the reaction is:
moles of HBr remaining = moles of HBr - moles of KOH
moles of HBr remaining = 0.025 mol - 0.015 mol = 0.010 mol

Next, calculate the total volume of the solution after the addition of KOH:
total volume = volume of HBr solution + volume of KOH solution
total volume = 0.050 L + 0.060 L = 0.110 L

Now, calculate the concentration of HBr remaining in the solution:
concentration of HBr = moles of HBr remaining / total volume
concentration of HBr = 0.010 mol / 0.110 L = 0.091 M

Finally, calculate the pH of the solution using the concentration of HBr remaining:
pH = -log[H+]
pH = -log(0.091) = 1.04

The calculated pH is approximately 1.04, so the closest option among the given choices is C) 1.7.

To solve this problem, you need to understand the concept of titration and the stoichiometry of the reaction between HBr and KOH.

First, let's determine the moles of HBr and KOH involved in the reaction:

Given:
- Volume of HBr solution = 50.0 mL = 0.050 L
- Concentration of HBr = 0.50 M

Moles of HBr = Volume × Concentration
= 0.050 L × 0.50 M
= 0.025 mol HBr

Now, let's determine the moles of KOH added during the titration:

Given:
- Volume of KOH solution added = 60.0 mL = 0.060 L
- Concentration of KOH = 0.25 M

Moles of KOH added = Volume × Concentration
= 0.060 L × 0.25 M
= 0.015 mol KOH

Next, we need to determine the limiting reagent. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product formed. In this case, we have an equal number of moles of HBr and KOH, so neither is in excess. Therefore, we need to calculate the moles of acid remaining after the reaction with KOH.

Moles of acid remaining = Moles of HBr - Moles of KOH
= 0.025 mol - 0.015 mol
= 0.010 mol HBr

To find the pH, we need to consider that HBr is a strong acid that completely dissociates in water, producing H+ ions. The concentration of H+ ions can be calculated by dividing the moles of HBr remaining by the total volume of the solution after the reaction.

Total volume of solution after reaction = Initial volume of HBr + Volume of KOH added
= 0.050 L + 0.060 L
= 0.110 L

Concentration of H+ ions = Moles of HBr remaining / Total volume of solution after reaction
= 0.010 mol / 0.110 L
= 0.091 M

Finally, we can calculate the pH using the formula:

pH = -log[H+]

pH = -log(0.091)
= 1.04 (rounded to two decimal places)

From the given options, the closest match to the calculated pH is 1.7 (option C). So, option C is the correct answer.