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MAVERICK [17]
3 years ago
8

8.38e-21 Q^2 -1.07e-23 Q^2 +3.15e+19 Solve for Q

Chemistry
1 answer:
GarryVolchara [31]3 years ago
4 0
8.38e -21Q^2 -1.07e -23Q^2 +3.15e +19
= 10.46e -44Q^2 + 19
44Q^2 = 10.46e + 19
Q^2 = 523/2200e + 19/44
Q1= ≈ -1.03828
Q2= ≈ 1.03828
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First, we need to calculate moles of hydrazoic acid NH3:

moles NH3 = molarity * volume 

                    = 0.15 m * 0.025 L

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moles NaOH = molarity * volume 

                       = 0.15 m * 0.015 L

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

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

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∴[NaOH] = moles NaOH / total volume 

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

Pka = -㏒Ka 

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PH = Pka + ㏒[salt/ basic]

PH = 4.7 +㏒[0.05625/0.09375]

∴ PH = 4.48 


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