Question
Write the reactions that occur for the benzoic acid/sodium benzoate buffer when you react it with HNO3 and KOH. How would I go about writing it? Thanks.
Answers
Let's call benzoic acid HBz and the benzoate ion then is Bz^-
With HNO3, the added acid (HNO3) reacts with the base, Bz^-.
Bz^- + H^+ ==> HBz
With KOH, the added base (KOH) reacts with the acid, HBz.
HBz + OH^- ==> Bz^- + H2O
With HNO3, the added acid (HNO3) reacts with the base, Bz^-.
Bz^- + H^+ ==> HBz
With KOH, the added base (KOH) reacts with the acid, HBz.
HBz + OH^- ==> Bz^- + H2O
Thank you so much!!!!
Related Questions
Given that the Ka of benzoic acid is 6.50 x 10-5, how would one prepare 0.500 L of a benzoic acid/so...
You are asked to prepare a pH=4.00 buffer starting from 1.50 L of 0.0200 \; M solution of benzoic a...
URGENT!!! How many grams of sodium benzoate (NaC6H5CO2) must be added to 500 mL of 0.15 M benzoic a...