Identifying base in a chemical equation

Question

According to the Brønsted Lowry principle, "base" is a proton acceptor. I've tried to apply this idea to every choice as you can see these red circles in each equation below, but they are all loses protons, aren't they? How could it be a base?

However, if I choose water in a blue circle, I will get the underlined substance as a base for both choice B and C instead. Also, if you have a look at other substances involve in equation B and C, they do support the assumption that they act as acids that involved in donating protons. Hence, both B and C are correct. But the solution said the correct answer is C.

Answer 1

I would write down the half equations for each reaction:

A

1. $\ce{ HNO3 + H+ -> H2NO3+ }$ (Base)
2. $\ce{H2SO4 -> HSO4- + H+}$ (Acid)

B

1. $\ce{ HSiO3- -> SiO2 + OH- }$ (Base)
2. $\ce{HCN -> CN- + H+ }$ (Acid)

C

1. $\ce{HNO2 -> NO2- + H+ }$ (Acid)
2. $\ce{HCO3- -> CO2 + OH- }$(Base)

D

1. $\ce{C6H5O- + H+ -> C6H5OH }$ (Base)
2. $\ce{CH2ClCO2H -> CH2ClCO2- + H+ }$(Acid)

Answer 2

There are multiple definitions for bases and acids.

The definition that most people thing of is the Brønsted-Lowry definition. An acid is a proton donor and a base is a proton acceptor. Which for bases implies a reaction of sort $$\ce{H2O + B- -> OH- + HB}$$

However there is also the Lewis definition of acids and bases. In this case acids act as electron acceptors and bases act as electron donors. (A way to remember this is that it is the inverse of Brønsted-Lowry! A base gives electrons instead of taking protons) An example of this would be $$\ce{H- + BH3 -> BH4-}$$ Here, hydride is acting as a lewis base and donating its electrons to borane, a very common lewis acid. Hydride is among the strongest, naturally observed lewis bases.

There are also Arrhenius acids and bases. Arrhenius acid's lose a proton whereas Arrhenius base's lose an $\ce{OH-}$. For example $$\ce{LiOH -> Li+ + OH-}$$

In the case of your problems. Equation C is an Arrhenius base. A quick way to check what is acting as a base is to write down the half reactions. Here, bicarbonate $\ce{HCO3-}$ is giving up a hydroxide ion to and nitrous acid is giving up a proton.