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A 20.0 ml sample of 0.150 m ethylamine is titrated with 0.050 m hcl. you may want to reference (page 805) section 17.4 while completing this problem. part a what is the ph after the addition of 5.0 ml of hcl? for ethylamine, pkb= 3.25.

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superman1987
First, we need to get moles of OH-:

moles of ethylamine = molarity * volume

                       = 0.15 M * 0.02 L 

                       = 0.003 mol


moles of H+ = molarity * volume

                     = 0.05 M * 0.005 L

                     = 0.00025 mol

when the total volume = 0.02 L + 0.005L 

                                        = 0.025 L

[H+] = moles / total volume

        =  0.00025 / 0.025 = 0.01 M

[ethaylamine] = moles / total volume

          = 0.003 / 0.025 = 0.12 M

when Pkb = 3.25  so we can get Pka from this formula:

Pka = 14 - Pkb 

       = 14 - 3.25

       = 10.75

by using H-H equation:

PH = Pka + ㏒[salt]/[acid]

       = 10.75 +㏒[0.12/0.01]

        = 11.8

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