Question

Good day,

I calculated The pH of a 0.425 M ammonia solution 11.44
Ammonia, NH3, is a weak base with a Kb value of 1.8×10−5.
What is the percent ionization of ammonia at this concentration?

Thanks

Answer 1

Percent ionization is calculated by estimating the fraction of ammonia dissociation from its initial concentration.

% ionization$=$ $[NH_3] _ { dissociation}\over [NH_3] _{initial}$$\over$ $*100$

$[NH_3]_{dissociation }=[NH_4^+]=[OH^-]$

$[NH_3]_{Initial}=0.425mol/L$

$pH+pOH=14$

$pH=11.44$

$pOH=2.56$

$pOH=-log [OH^-]$

$2.56=-log [OH^-]$

$10^{-(2.56)}=[OH^-]$

$[OH^-]=10^{-(2.56)}=0.00275mol/L$ percent ionization$=$ $($ $0.00275\over 0.425$ $)$ $*100$

.$=0.647$

Percentage ionization is thus $0.647$