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poizon [28]
3 years ago
10

You are performing a titration of a triprotic acid, when you spill water on your lab notebook. you can read that: pka 1 = 1.40,

pka 3 = 9.80. you have determined experimentally that the ph at the first equivalence point is 3.35, and the ph at the second equivalence point is 7.55. what is pka 2 for this acid?
Chemistry
1 answer:
eimsori [14]3 years ago
5 0
According to the PH formula:
PH= Pka +㏒ [strong base/weak acid]
when we have PH at the first equivalence =3.35 and the Pka1 = 1.4
So, by substitution, we can get the value of ㏒[strong base / weak acid]
3.35 = 1.4 + ㏒[strong base/ weak acid]
∴㏒[strong base/weak acid] = 3.35-1.4 = 1.95 
to get the Pka2 we will substitute with the value of ㏒[strong base/ weak acid] and the value of PH of the second equivalence point
∴Pk2 = PH2 - ㏒[strong base/ weak acid]
          = 7.55 - 1.95 = 5.6 
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wlad13 [49]

Answer:

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where p₀= initial pressure , ρ=density , g = gravity

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ΔP= ρ*g*Δh → ΔP/Δh =  ρ*g = ( 15 psi - 3 psi) /( 8.6 m - 5.5 m)  = 12/3.1 psi/m

if the liquid level is converted linearly

P = P₁ + ΔP/Δh*(h -h₁)

therefore choosing  P₁ = 3 psi and h₁= 5.5 m , for h=7.2 m

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4.7 psi = 3 psi  + 12/3.1 psi/m *(h -5.5 m)

h = (4.7 psi - 3 psi)/ (12/3.1 psi/m) + 5.5 m = 5.94 m

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