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
The statement which describes how NO2- reacts in this equilibrium:
<span>H2SO3(aq) + NO2-(aq) HSO3-(aq) + HNO2(aq
is the second option - </span><span>B. as a Brønsted-Lowry base by accepting a proton.
</span>This is because bases take proton H+ in order to become HNO2.
<h3>A. an acidic............</h3>