Question

Selenic acid (H2SeO4) is diprotic. It can react with sodium hydroxide in steps.

a) Write equations (in molecular form) for the two steps of the reaction between H2SeO4 and NaOH. [You do not have to indicate physical states of reactants and products here.]

b) Name the two salts that form in the above reactions.

c) Which salt(s) and how many moles (of each) will form from 0.50 mol of selenic acid and 0.15 mol of sodium hydroxide?

d) Which salt(s) and how many moles (of each) will form from 0.50 mol of selenic acid with 0.85 mol of sodium hydroxide?

e) Which salt(s) and how many moles (of each) will form from 0.50 mol of selenic acid with 1.45 mol of sodium hydroxide?

Answer 1

a)

two step reaction of H2SeO4 + NaOH:

H2SeO4 + NaOH --> H2O + Na+ + HSeO4-

HSeO4-+ NaOH --> H2O + Na+ + SeO4-2

add all:

H2SeO4 + 2NaOH --> 2H2O + 2Na+ + SeO4-2

b)

salts formed:

NaHSeO4 = Sodium biselenate (bi stands for Hydrogen, NOT double amount of Na ions)

Na2SeO4 = Sodium selenate

c)

0.5 mol of selenic acid + 0.15 mol of NAOH

not enouth to neutralize 1st proton so

H2SeO4 + NaOH --> H2O + Na+ + HSeO4-

H2SeO4 and HSeO4- ar eprestn

mol of H2SeO4 left = 0.50-0.15 = 0.35 mol

mol of HSeO4formed = 0 + 0.15 = 0.15 mol

d)

mol of H2SeO4 left = 0.5-0.5 = 0

HSeO4 - formed = 0.5-(0.85-0.50) = 0.15

left

e)

mol of H2SeO4 left = 0.5-0.5 = 0

HSeO4 - formed = 0.5-(1.45-0.50) = 0

SeO4-2 formed = 0.5 mol formed