Question

In a titratiom 0.140 M Hbr is titrated by 0.100 M NaOH. If 16.00 mL of the NaOH titrant is added to 20.00 mL of the HBr solution (followed by mixing). What is the pH of the resulting solution?  

A. 2.30
B. 1.48
C. 4.60
D. 1.84
E. 1.20

Answer 1

Number of mmol of HBr is , n = Molarity x volume in mL

                                              = 0.140 M x 20.00 mL

                                             = 2.8 mmol

Number of mmol of NaOH is , n' = Molarity x volume in mL

                                              = 0.100 M x 16.00 mL

                                             = 1.6 mmol

       NaOH + HBr NaBr + H₂O

1 mole of NaOH reacts with 1 mole of HBr

                          OR

1 mmole of NaOH reacts with 1 mmole of HBr

1.6 mmole of NaOH reacts with 1.6 mmole of HBr

So 2.8-1.6=1.2 mmol of HBr left unreacted in the solution which imparts acidic nature to the solution.

Concentration of HBr in the solution ,

                                                    

HBr H+ + Br-

1 mole of HBr produces 1 mole of H⁺

So the concentration of H⁺ = [H⁺ ] = concentration of HBr = 0.033 M

pH = - log [H⁺ ]

     = - log 0.033

     = 1.48

So option (B) is correct