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A solution of volume 70.0 mL contains 16.0 mmolHCHO2 and 8.50 mmol NaCHO2. Part A What...

A solution of volume 70.0 mL contains 16.0 mmolHCHO2 and 8.50 mmol NaCHO2.

Part A

What is the pH of this solution?

Express your answer using two decimal places.

Part B

If 0.25 mmol Ba(OH)2 is added to the solution, what will be the pH?

Express your answer using two decimal places.

Part C

If 1.10 mL of 12 M HCl is added to the original solution, what will be the pH?

Express your answer using two decimal places.

Solutions

Expert Solution

Pka of HCO2H = 3.75

no of mmoles of HCO2Na = 8.5mmoles

no of mmoles of HCO2H = 16 mmoles

PH = PKa + log[HCO2Na]/[HCO2H]

      = 3.75 + log8.5/16

      = 3.75-0.2747   = 3.4753

part-B

no of mmoles of HCO2Na after addition of 0.25 mmoles of Ba(OH)2 = 8.5+0.25   = 8.75mmoles

no of mmoles of HCO2H after addition of 0.25 mmoles of Ba(OH)2   = 16-0.25      = 15.75 mmoles

    PH = Pka + log[HCO2Na]/[HCO2H]

          = 3.75 + log8.75/15.75

          = 3.75-0.2552   = 3.4948>>>answer

part-C

no of mmoles of HCl = molarity * volume in ml

                                   = 12*1.1   = 13.2mmoles

The excess of HCl total converted HCO2Na to HCO2H

no of mmoles of excess = 13.2-8.5   = 4.7 mmoles

conc of HCl   = no of mmole of HCl/total volume in ml   = 4.7/70.1 = 0.067M

[H^+]   = 0.067M

PH = -log[H^+]

       = -log0.067

       = 1.1739 >>>>>answer

queen_honey_blossom answered 2 years ago
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