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 Calculate amount of ammonium chloride required to dissolve in 500ml water to make pH = 4.5 [ Kb(NH3) = 1.8 x 10-5 ]??????

Please I need a answer.......

Dear Student,

@ wannachat,

Good effort! You have applied the correct formula to solve this question.

But it is not complete, I would like to solve it further to get the answer:

x = amount of NH4Cl (ammonium chloride) dissolved in 500 ml of water

 NH4Cl is a salt of strong acid and weak base so

pH =  1/2(pKw - log C- pKb )

4.5 =  1/2[14 - log C + log (1.8×10-5)]

9 =  14 - log C - 4.74

 log C = 0.26

 C = x/26.25

log (x/26.25) = 0.26

log x - log 26.25 = 0.26

log x = 0.26 + 1.4191 = 1.6791

x = Antilog (1.6791)

x = 47.76 

Amount of NH4Cl dissolved =47.76 g 

Hope it helps!

  • 17
View Full Answer

however much NH3 u put it wont be 4.5 coz NH3 is a base....... its pH will be more than 7 only..........

  • 2

no wait..............sry i didn't read the ques properly............

  • 3

Let amt of NH3Cl be x

.: M = 1000/500 *  x/52.5 =  x / 26.25

this is molar conc of NH3Cl

pKb=  - log Kb =  - log1.8 + 5 log10

now formula for calculating pH of salt made from weak base and strong acid is :-

pH = 7 - 1/2 ( pKb + log conc )

4.5  =  7 - 1/2 ( - log1.8 +  5 log10 +  log x / 26.25 )

4.5  =  7 - 1/2 ( - log1.8 +  5 log10 +  log x -  log 26.25 )

  • 4

plz. do this harissingly long calculation on your own............ n if u want me to do it give thumbs up/ down n leave a reply .... i will do it.

  • -4

 Thanks, I tried and got the answer frm a diff mathod...

[H+]= 10-4.5 = 3.162 x 10-5 M

NH4+ -----> NH3 + H+

C(1-h)         Ch         Ch

Kh = Ch2 (h<<1)

Kh = Kw / Kb = 10-14 / 1.8 x 10-5 = 5.5 x 10-10

:. h = Kh / Ch = 3.162 x 10-5 / 1.74 x 10-5 = 1.8mol / L

500 ml of H2O contains 1.8 / 2 = 0.9 moles

Mass in gms = 0.9 x 53.5 = 48.15 gm

  • 9

ThanQ!!!!!!!!!!!!

  • 0
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