Question

In a titration, 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?

Answer 1

HBr + NaOH ---> NaBr +H20

                 HBr               +        NaOH --->       NaBr        +      H20

Initial                                 0.0028 moles               0.0016 moles

number of moles of( HBr) ==> 0.140 = (n*1000)/ 20==> 0.0028 moles

number of moles of( NaOH) ==>0.1 = (n*1000)/ 16==> 0.0016 moles

[H+]=[HBr+]= 0.0028mole*1000/36 ml

==>0.0778M==>[H+]=0.0778M

pH= -log[H+]= -log[0.0778]=1.109

therefore pH=1.109