Phosphoric acid has 3 pKa values (pKa1=2.1, pKa2=6.9, pKa3= 12.4) and after 3 ionization it gives 3 types of ions at different pKa values:
H₃PO₄(aq) + H₂O(l) ⇌ H₃O⁺(aq) + H₂PO₄⁻ (aq) pKₐ₁
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</span>H₂PO₄⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + HPO₄²⁻ (aq) pKₐ₂
HPO₄²⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + PO₄³⁻ (aq) pKₐ₃
The last equilibrium is associated with the highest pKa value (12.4) of phosphoric acid. There the last OH group will lose its hydrogen and hydrogen phosphate ion (HPO₄²⁻) turns into phosphate ion (PO₄³⁻).
Explanation:
The given data is as follows.
= 0.483,
= 0.173 M,
= 0.433 M,
= 0.306 M,
= 9.0 atm
According to the ideal gas equation, PV = nRT
or, P =
Also, we know that
Density = 
So, P = MRT
and, M = 
= 
= 
= 0.368 mol/L
Now, we will calculate the cell potential as follows.
E = ![E^{o} - \frac{0.0591}{n} log \frac{[Co^{2+}]^{2}[Cl_{2}]}{[Co^{3+}][Cl^{-}]^{2}}](https://tex.z-dn.net/?f=E%5E%7Bo%7D%20-%20%5Cfrac%7B0.0591%7D%7Bn%7D%20log%20%5Cfrac%7B%5BCo%5E%7B2%2B%7D%5D%5E%7B2%7D%5BCl_%7B2%7D%5D%7D%7B%5BCo%5E%7B3%2B%7D%5D%5BCl%5E%7B-%7D%5D%5E%7B2%7D%7D)
= 
= 
= 
= 0.483 - 0.0185
= 0.4645 V
Thus, we can conclude that the cell potential of given cell at
is 0.4645 V.
Answer:
C. A question and opinions
Explanation:
The octet rule is that atoms must have 8 valence electrons to have an octet. Usually, when atoms have an octet, they tend not to bond with any other atoms because they are already stable and do not need to bond anymore.