A 0.45 M solution of Ammonium Chloride has an acidic pH . A. Write a net ionic e

Question

A 0.45 M solution of Ammonium Chloride has an acidic pH . A. Write a net ionic equation . B- for the reaction in part a, write the equilibrium constant from a consideration of Kb and Kw. C- calculate the h3o^+ for a 0.45 M nh4cl solution . D- calculate the percent of hydrolysis of 0.45 M NH4Cl.

Explanation / Answer

Ok first we need to know what happens when you put ammonium chloride in water, the overall equation is this: NH4Cl >> NH4+ +Cl- remember that this reaction does not go into completion because NH4Cl is a weak acid; the ammonium ion is going to reac with water to give you ammonia and H3O+ NH4+ + H2O>> NH3 + H3O+ Let's do a reaction table, to see how we pass from initial, change and equilibrium concentrations: NH4Cl >> NH4+ +Cl- Initial 0.10M 0 0 change -0.10 +0.10 +0.10 equil. 0 0.10M 0.10M now we know that the concentration of NH4+ is 0.10M; we used another equil. table using the second equation: NH4+ + H2O>> NH3 + H3O+ initial 0.10M 0 0 change -x +X +x equil 0.10-x x x Q for this reaction is : [ NH3] [H3O+]/[NH4+] we changed Kb for Ka using the relationship: Ka= Kw/Kb = Ka= 1.0x10^-14/1.8x10^-5 Ka= 5.6x10^-10 5.6x10^-10= [x][x]/[0.10-x] because Ka is so low we can forget about x: 5.6x10^-10= (x}2/(0.10)= (x}2= 5.6x10^-11= x= 7.48x10^-6 [H3O+] =7.48x10^-6 NH4+ + H2O ---> NH3 + H3O+ [H3O+] = square root (Kw/Kb*Ca) = 7.45*10^-6 M

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