What is the ph of 0.10 m nh4cl(aq). the kb of nh3 is 1.8 x 10-5 8.87 11.13 5.13 2.87

Hey there!:

ka = kw/ kb

Ka = 1.0*10⁻¹⁴ / 1.8* 10⁻⁵

Ka = 5.56*10⁻¹⁰

Now Cl⁻ is spectator ion

NH⁴⁺ --->  NH3 + H⁺

Kb =  [NH⁴⁺ / [NH3] [H⁺]

at equilibrium  the concentration of products is taken as x each  and that of ammonium  ion will be  ( 0.10 -x )

5.56*10⁻¹⁰ = x² / [0.10 -x]

x =  7.54* 10⁻⁶ M

pH = - log [ H⁺ ]

pH = - log [ 7.54*10⁻⁶ ]

pH =  5.13

Answer C

Hope that helps!

answer: mercury

explanation: