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?
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?
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?
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 = ?
0.0000196 = M?
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
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?
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.