With the Ka= 1.8x10^-5, what is the Ph?

Question

CH3COOH(100 mL OF 1.0M) + OH(50 mL OF 1.0M)<--->H3O(aq) + CH3COONa(aq)

Answer 1

when ever 1/2 of the acid has been neutrlized into its salt, pH = PKa  

when K = 1.8e-5, pKa = 4.74

your answer is 4.74

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CH3COOH(0.100 L OF 1.0mol / litre = 0.100 moles acid)

OH(0.050 L OF 1.0mol/Litre = 0.050 moles OH-)

these react producing 0.050 moles of CH3COONa,

while 0.050 moles of acetic acid remains

CH3COOH + OH- &lt;---&gt;H3O+  &amp; CH3COO-

K = [H3O+] [CH3COO-] / [CH3COOH]

1.8e-5 = [H3O+] [0.050] / [0.050]

H+ = 1.8e-5

pH = 4.74