Question

suppose you added 26.0 ml of 0.01 M NaOH to the solution, what is the molarity of excess NaOH in the mixture? what is the ph of this solution?

Answer 1

Mollie, I know all of the tutors on this site are good but we don't have ESP or any of those extraordinary powers. In other words, what solution are you talking about?

Answer 2

I am sorry I got ahead of myself! It would be kind of important! We have 26.0 ml 0.01 M NaOH and we are adding it to 25.0 ml 0.01 M HCL. what is the molarity of excess NaOH in the mixture?

Answer 3

Yes, I would say the missing information is important (at least to the problem).
I like millimoles (you can change to mols if you like by dividing millimoles by 1000). I like millimoles because it keeps all of those zeros on the left from showing up.
millimols NaOH = 26.0 x 0.01 = 0.260
millimols HCl = 25.0 x 0.01 = 0.250
0.260-0.250= 0.01 millimols excess NaOH

Then M = mols/L (and since I'm working in millimoles, M = mmols/mL = 0.01 mmol/(26.0 + 25.0)mL = ?

Answer 4

0.0000196 = M?

Answer 5

I don't think so. Count those zeros. I think you have one too many. Now you know what I don't like about those pesky things.
0.01 mmols/51 mL = 1.96E-4M

0.026L x 0.01M = 0.00026 mols NaOH
0.025L x 0.01M = 0.00025 mols HCl
0.00001 mols/0.051 L = 1.96E-4

Or you can let the calculator keep up with all of those zeros this way.
(2.6E-4 - 3.5E-4)/0.051 = 1.96E-4 M

Answer 6

oh okay! I see what I did wrong. I forgot the 0 out of the 0.051. I had 0.01/0.51 and that gave me too many zeros. So then how would you calculate the pH of the solution?

Answer 7

This is an excess of OH^-; therefore, (OH^-)= 1.96E-4M
Then pOH = -log(OH^-) = ?
Then pH + pOH = pKw = 14.
You know pKw and pOH, solve for pH.