Calculate the pH of the solution that results when 15.0 mL of 0.1630 M benzoic a

Question

Calculate the pH of the solution that results when 15.0 mL of 0.1630 M benzoic acid is (This problem requires values in your textbook's specific appendices, which you can access through the OWLv2 MindTap Reader. You should not use the OWLv2 References' Tables to answer this question as the values will not match.) diluted to 40.0 mL with distilled water. pH = mixed with 25.0 mL of 0.0978 M NaOH solution. pH = mixed with 25.0 mL of 0.160 M NaOH solution. pH = mixed with 25.0 mL of 0.160 M sodium benzoate solution. pH

Explanation / Answer

a) FROM DILUTION LAW

M1V1 = M2V2

(15*0.163) = (40*M2)

M2 = 0.061 M

pka of benzoicacid = 4.204

pH = 1/2(pka-logC)

    = 1/2(4.204-log0.061)

   = 2.71

B) nO of mol of benzoicacid = 15/1000*0.163 = 0.00244 mol

    nO of mol of NaOH = 25/1000*0.0978 = 0.00244 mol

at equivalencepoint

concentration of salt = 0.00244/0.04 = 0.061 M

pH = 7+1/2(pka+logC)

     = 7+1/2(4.204+log0.061)

    = 8.5

c) nO of mol of benzoicacid = 15/1000*0.163 = 0.00244 mol

   nO of mol of NaOH = 25/1000*0.16 = 0.004 mol

concentration of excess NaOH = (0.004-0.00244)/0.04 = 0.039 M

pH =14-(-log(0.039)) = 12.6

d) nO of mol of benzoicacid = 15/1000*0.163 = 0.00244 mol

   nO of mol of sodium benzoate = 25/1000*0.16 = 0.004 mol

   pH = pka + log(salt/acid)

      = 4.204+log(0.004/0.00244)

     = 4.42

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