9.50mL of .100M NaOH + 8.00mL of .100M of HC2H3O2. Find the pH.

Okey this is what I did.

8.00-9.50=1.5mL or .0015L

.0015L x .1000 = .00015 moles

.00015/(total volume) = .0085714 M

-log(.0085714)= 2.0669=pOH
14-2.0669 = 11.933

But I'm getting the feeling that this is wrong. Thanks.

That looks ok to me except you didn't label 11.93 s the pH. If you find differently, please let me know.

THANKS DR.BOB

To find the pH of a solution resulting from the reaction between NaOH and HC2H3O2, you need to consider the stoichiometry of the reaction and the dissociation of the weak acid, HC2H3O2.

First, write the balanced chemical equation for the reaction:
NaOH + HC2H3O2 -> H2O + NaC2H3O2

In this reaction, one mole of NaOH reacts with one mole of HC2H3O2.

Next, calculate the moles of NaOH and HC2H3O2:
9.50 mL of 0.100 M NaOH = (0.0095 L) x (0.100 mol/L) = 0.00095 moles NaOH
8.00 mL of 0.100 M HC2H3O2 = (0.008 L) x (0.100 mol/L) = 0.0008 moles HC2H3O2

Based on the stoichiometry of the reaction, since NaOH and HC2H3O2 react in a 1:1 ratio, the limiting reagent is HC2H3O2 since it has fewer moles.

Now, calculate the moles of HC2H3O2 remaining:
0.0008 moles - 0.0008 moles = 0 moles HC2H3O2

Since HC2H3O2 is a weak acid, it will partially dissociate in solution. The final pH of the solution will depend on the concentration of the remaining HC2H3O2.

To calculate the remaining concentration of HC2H3O2, use the formula:
[H+] = Ka * [HC2H3O2] / [NaC2H3O2]

The Ka value for acetic acid (HC2H3O2) is 1.8 x 10^-5.

Assuming that the volumes of the solutions are additive, the total volume is:
9.50 mL + 8.00 mL = 17.50 mL = 0.0175 L

Now calculate the concentration of HC2H3O2:
[HC2H3O2] = 0.0008 moles / 0.0175 L = 0.0457 M

Substituting the values into the equation for [H+]:
[H+] = (1.8 x 10^-5) * (0.0457 M) / (0.0008 M) = 0.001017 M

Finally, calculate the pH using the equation:
pH = -log [H+]

pH = -log (0.001017) = 2.99

Therefore, the pH of the solution resulting from the reaction is approximately 2.99.