A25.0 ml sample of 0.150 m hydrazoic acid is titrated with a 0.150 m naoh solution. what is the ph after 15.0 ml of the sodium hydroxide solution is added? the ka of hydrazoic acid is 1.9×10-5.

First, we need to calculate moles of hydrazoic acid NH3:

moles NH3 = molarity * volume 

                    = 0.15 m * 0.025 L

                   =  0.00375 moles

moles NaOH = molarity * volume 

                       = 0.15 m * 0.015 L

                       = 0.00225 moles 

after that we shoul get the total volume = 0.025L + 0.015L

                                                                   = 0.04 L

So we can get the concentration of NH3 & NaOH by:

∴[NH3] = moles NH3 / total volume 

           = 0.00375 moles / 0.04 L

           = 0.09375 M

∴[NaOH] = moles NaOH / total volume 

                = 0.00225 moles / 0.04 L

                = 0.05625 M

then, when we have the value of Ka of NH3 so we can get the Pka value from:

Pka = -㏒Ka 

       = - ㏒ 1.9 x10^-5

      = 4.7 

finally, by using H-H equation we can get PH:

PH = Pka + ㏒[salt/ basic]

PH = 4.7 +㏒[0.05625/0.09375]

∴ PH = 4.48 

Cis the

No is the answer to the question