If 30.0 mL of 4.8 10-2 M HBr is added to 21.5 mL of 3.6 10-2 M NaOH, what is the pH of the solution?

HBr + NaOH ==> NaBr + H2O

moles HBr = M x L
moles NaOH = M x L
There can be three scenarios:
a. If moles HBr exactly equals moles NaOH, then you have the salt, NaBr, in water and the pH = 7 because neither Na^+ nor Br^- hydrolyze.

b. If more moles HBr than NaOH are present, there will be excess HBr and the pH will be determine by the molarity of the HBr (it is a strong acid.)

c. If more moles NaOH than HBr are present, there will be an excess of NaOH and the pH will be determined by the molarity of the NaOH (it is a strong base).

Post your work if you get stuck.

I did:

(4.8e-2*30.0mL)/1000mL=.0014
(3.6e-2*21.5mL)/1000mL=7.74e-4
Then I added the two to get .005474 then divided that by the total L to get .106
I then took the -log(.106) to get .975, but that's the wrong answer

Why in the world did you add them? And where did the 1000 mL come from? I don't see that in my instructions. Nor do I see anything about taking the log in my instructions.

so (4.8e-2*30.0mL)=1.44

(3.6e-2*21.5mL)=.774
now what do I do

Great. Now look at the equation.This is an acid/base reaction. Acids and bases react to give a salt + water.

HBr + NaOH ==> NaBr + H2O

I can't write these across the board because it ignores the spaces. Let me use periods (like this .......) to make it space correctly. Just ignore the periods.
HBr + NaOH = NaBr + H2O
1.44...774....0.....0 initially
-.774..-.774..+.774..+.774 change
.666....0.......774.......final

The HBr is in excess, the concn is now .666/51.5 mL = ??
Now you can do the pH = -log(H^+) thing)
I just figured out where your 1000 came from. You were changing mL to L. I didn't do that above; I left everything in mL and concn at the end is millimoles/mL. I think the pH is about 1.8 (a guess) or so but check me out on that.

ohh..I understand now, I forgot all about the ice table...I got 1.8 as well and that is the correct answer...Thanks!

To find the pH of the solution, we first need to determine the number of moles of HBr and NaOH, then calculate the moles of the excess reactant, and finally determine the concentration of the remaining reactant.

Step 1: Calculate the moles of HBr
Moles of HBr = volume (L) × concentration (M)

Given:
Volume of HBr = 30.0 mL = 0.0300 L
Concentration of HBr = 4.8 × 10^(-2) M

Moles of HBr = 0.0300 L × 4.8 × 10^(-2) M = 0.00144 moles

Step 2: Calculate the moles of NaOH
Moles of NaOH = volume (L) × concentration (M)

Given:
Volume of NaOH = 21.5 mL = 0.0215 L
Concentration of NaOH = 3.6 × 10^(-2) M

Moles of NaOH = 0.0215 L × 3.6 × 10^(-2) M = 0.000774 moles

Step 3: Determine the Excess Reactant
To determine which reactant is in excess, compare the moles of HBr and NaOH. The reactant with the smaller mole value is the excess reactant.

In this case, NaOH has the smaller mole value, so it is the excess reactant.

Step 4: Calculate the moles of the remaining reactant
To find the moles of HBr that reacted, subtract the moles of NaOH from the moles of HBr.

Moles of HBr reacted = Moles of HBr - Moles of NaOH
= 0.00144 moles - 0.000774 moles
= 0.000666 moles

Step 5: Calculate the concentration of the remaining HBr
To find the concentration of the remaining HBr, divide the moles of HBr reacted by the total volume of the solution.

Total volume of the solution = volume of HBr + volume of NaOH
= 0.0300 L + 0.0215 L
= 0.0515 L

Concentration of the remaining HBr = Moles of HBr reacted / Total volume of solution
= 0.000666 moles / 0.0515 L
= 1.29 × 10^(-2) M

Step 6: Calculate the pH of the solution
The pH of the solution can be calculated using the concentration of HBr.
pH = -log10(concentration of HBr)

pH = -log10(1.29 × 10^(-2))
= 1.89

Therefore, the pH of the solution is 1.89.