A 25.0 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