# Identify reactants and products as acid, base or neither

Identify each reactant and product in the following chemcial reactions as a acid, a base, or neither. Arrange the species in each reaction as conjugate acid-base pairs.

a)  NH3 + HI = NH4+ + I

b) CH3COOH + NO2 = NH02 + CH3COO

2 Classify the following samples from most acidic to most basic:

a) blood (ph 7.4)
b) A 1 M solution of hydorchloric acid
c) tomatoes (ph 4.5)
d) milk (ph 6.5)
e) a solution of sodium hydroxide at 0.100 mol.dm3
f) a solution b that has [H3O+] = 8.2 X 10-2 mol.dm3
g) a neutral solution of water

If someone could help, I would appreciate it a lot thank you very much!!

Steven

For every acid there is a conjugate base, formed by loss of a proton
For every base there is a conjugate acid, formed by gain of a proton
In your first example, NH3 + HI --> NH4 +  +  I-
NH3 is the base (because in the reaction it accepts a proton from the acid),
HI is the acid (because it donates a proton, ammonium ion is the conjugate acid while I- is the conjugate base.
In the second eg, CH3COOH is the acid, NO2 is the base, while CH3COO-

while CH3COO- is the conjugate base i got it thank you very much

previous post incomplete. Use this
For every acid there is a conjugate base, formed by loss of a proton
For every base there is a conjugate acid, formed by gain of a proton
In your first example, NH3 + HI --> NH4 +  +  I-
NH3 is the base (because in the reaction it accepts a proton from the acid),
HI is the acid (because it donates a proton, ammonium ion is the conjugate acid while I- is the conjugate base.
In the second eg, CH3COOH is the acid, NO2 is the base, while CH3COO- is the conjugate base (because it can accept a proton) and HNO2+ is the conjugate acid.

In the second question, the lower the pH, the more acidic. The higher the pH, the less acidic.
You will need to calculate the pH values where they are not given e.g. 1M HCl has [H+] = 10^1 mol l^1. pH = -log[H+] = 1 - so this is very acidic. I assume you know [H+][OH-] = 1 x 10^-14 mol^2 L^-2.
For the NaOH, [OH-] = 10^-1 mol/l so [H+] = 10^-13 mol/l so the pH = 13.
For unknown solution b, just put [H3O+] into the pH equation and calculate.