The acid constant of acetic acid (HOAc) is Ka = 1.85 x 10-5. Calculate pH of the

Question

The acid constant of acetic acid (HOAc) is Ka = 1.85 x 10-5. Calculate pH of the solution prepared by dissolving 16.2 mmol of NaOAc in 250 mL of 0.100 M HOAc adding 125 mL of 0.0500 M NaOH to 125 mL of 0.175 M HOAc adding 50.0 mL of 0.120 M HC1 to 200 mL 0.0420 M NaOAc

Explanation / Answer

a) pH = pka + log[NaOAc][HOAc] , [NaOAc] = 0.0162/0.25 =0.0648 M pH = -log( 1.85 x10^ -5) + log( 0.0648/0.1) = 4.544 , b) HOAC ---> OAc- + H+ , [HOAc] =0.175-x , [H+][OAc-] = x , Ka= 1.85 x10^ -5 = x^2/(0.175-x) , x = 0.001788 = [H+] , H+ moles = 0.001788 x0.125 = 0.0002235 , OH- moles = 0.125 x0.05 =0.00625 , net OH- moles = 0.00625-0.0002235=0.006026 , [OH-] = 0.006026 x1000/250 =0.024 , pOH = -log(0.024) = 1.162 , pH = 14-1.162=12.838 , c) Kb for NaOAc = Kw/Ka = 10^ -14/1.85 x10^ -5 = 5.4 x10^ -10 , NaOAc + H2O ---> HAOC + H2O , [NaOAc] = 0.042-x , [OH-] =[HOAc] = x , Kb = 5.4 x10^ -10 = x^2/(0.042-x) , x = 4.76 x10^ -6 =[OH-] , OH- moles = 4.76x10^ -6 x 0.2 =9.525 x10^ -7 , H+ moles = 0.05 x0.12 =0.006, net[ H+ ]= (0.006-9.525 x10^ -7)/0.25 = 0.0234 , pH = -log(0.0234) = 1.63

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