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boric acid (monoprotic acid of type HA with ka=5.0x10-5) is titrated against .200m naoh(aq) a) calculate...

boric acid (monoprotic acid of type HA with ka=5.0x10-5) is titrated against .200m naoh(aq)

a) calculate the ph after 20.0ml of .200 m Noah (aq) has been added to 50.0ml of .200m boric acid.

b) calculate the ph at the equivalence point

please show steps thank you

Solutions

Expert Solution

millimoles of boric acid = 50 x 0.2 = 10

millimoles of NaOH = 20 x 0.2 = 4

H3BO3 +   NaOH --------------------> NaH2BO3 + H2O

10                 4                                       0                 0

6                  0                                        4                 4

here salt and acid remained. so it forms acidic buffer.

pH = pKa + log [salt / acid]

      = 4.3 + log [4 / 6]

      = 4.12

pH = 4.12

b)

At equivalence point only salt remained.

salt millimoles = 10

salt concentration = 10 / (50+50)

                              = 0.1

pH = 7+ 1/2 [pKa + log C]

      = 7 + 1/2 [4.3 + log 0.1]

       = 8.65

pH = 8.65

queen_honey_blossom answered 2 months ago
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