chem--need help, please!!!

Ka for benzoic acid, C6H5COOH, 6.5x10^-5. Calculate the pH of solution after addition of 10.0, 20.0, 30.0, and 40.0 mL of 0.10 M NaOH to 40.0 mL of 0.10 M Benzoic acid.
PLEASE CHECK MY ANSWER!!!!!

My answer is:

Moles acid = 0.040 L x 0.10 M = 0.0040
Moles base = 0.010 L x 0.10 M = 0.0010
C6H5COOH + OH- >> C6H5COO- + H2O
moles acid = 0.0040 -0.0010 = 0.0030
moles salt = 0.0010
total volume = 10 + 40 = 50 mL = 0.050 L
concentration acid = 0.0030 / 0.050 =0.060 M
concentration salt = 0.0010 / 0.050 =0.020 M
pKa = 4.19
pH = 4.19 + log 0.020 / 0.060 =3.71

work in the same way I get answers for the other questions.

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  1. Looks good to me. Thanks for showing your work. I was about to pass it by until I saw you wanted an answer check.

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  2. thank you very much, DrBob222!

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