What is #[NH_4^+]# in a solution that is 0.0200 M #NH_3# and .0100 M #KOH#?

1 Answer
Jun 20, 2016

Answer:

#[NH_4^+]=3.54xx10^(-5)" ""mol/l"#

Explanation:

Ammonia is a weak base and partially ionises in water:

#NH_3+H_2OrightleftharpoonsNH_4^++OH^-#

For which:

#K_b=([NH_4^+][OH^-])/([NH_3])=1.77xx10^(-5)" ""mol/l"#

#:.[NH_4^+]=K_bxx[[NH_3]]/[[OH^-]]#

The concentrations given are at equilibrium so:

#[NH_4^+]=1.77xx10^(-5)xx0.02/0.01=3.54xx10^(-5)" ""mol/l"#