Oxygen carbon and hydrogen
The dissociation of formic acid is:
The acid dissociation constant of formic acid, 
is:
![k_a = \frac{[HCOO^{-}] [H^{+}]}{HCOOH}](https://tex.z-dn.net/?f=%20k_a%20%3D%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%20%20%5BH%5E%7B%2B%7D%5D%7D%7BHCOOH%7D%20%20%20%20%20)
Rearranging the equation:
![\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%7D%7B%5BHCOOH%5D%7D%20%3D%20%5Cfrac%7Bk_a%7D%7B%5BH_%2B%5D%7D%20)
pH = 2.75
![pH = -log[H^{+}]](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%5BH%5E%7B%2B%7D%5D%20)
![[H^{+}]= 10^{-2.75} = 1.78 \times 10^{-3}](https://tex.z-dn.net/?f=%20%5BH%5E%7B%2B%7D%5D%3D%2010%5E%7B-2.75%7D%20%3D%201.78%20%5Ctimes%2010%5E%7B-3%7D%20)
Substituting the values in the equation:
![\frac{[HCOO^{-}]}{[HCOOH]} = \frac{1.78\times 10^{-4}}{1.78\times 10^{-3}}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%7D%7B%5BHCOOH%5D%7D%20%3D%20%5Cfrac%7B1.78%5Ctimes%2010%5E%7B-4%7D%7D%7B1.78%5Ctimes%2010%5E%7B-3%7D%7D%20%20%20)
Hence, the ratio is 
.
Answer:
74.4 ml
Explanation:
C₆H₈O₇(aq) + 3NaHCO₃(s) => Na₃C₆H₅O₃(aq + 3CO₂(g) + 3H₂O(l)
Given 15g = 15g/84g/mol = 0.1786mole Sodium Bicarbonate
From equation stoichiometry 3moles NaHCO₃ is needed for each mole citric acid or, moles of citric acid needed is 1/3 of moles sodium bicarbonate used.
Therefore, for complete reaction of 0.1786 mole NaHCO₃ one would need 1/3 of 0.1786 mole citric acid or 0.0595 mole H-citrate.
The question is now what volume of 0.8M H-citrate solution would contain 0.0595mole of the H-citrate? This can be determined from the equation defining molarity. That is => Molarity = moles solute / Liters of solution
=> Volume (Liters) = moles citric acid / Molarity of citric acid solution
=> Volume needed in liters = 0.0.0595 mole/0.80M = 0.0744 Liters or 74.4 ml
Answer:
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