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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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Boyle's Equation:     \frac{P_{1} }{T_{1} }    =   \frac{P_{2} }{T_{2} }
Since the initial temperature (T₁) is 25 C, the final temperature is 50 C (T₂) and the initial pressure (P₁) is 103 kPa, then we can substitute these into the equation to find the final pressure (P₂).
                      \frac{P_{1} }{T_{1} }    =   \frac{P_{2} }{T_{2} }∴ by substituting the known values,                 ⇒       (103 kPa) ÷ (25 °C)  =  (P₂) ÷ (50 °C)
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The structure of XeH_4 is shown below.

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<u />

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