What is the pH of 10ml of 0.1M NH3, with 5ml of 0.1M of HCL. kb=1.8*10^-5

Expert Solution

no of moles of NH3 = molarity * volume in L

= 0.1*0.01 = 0.001 moles

no of moles of HCl = molarity * volume in L

= 0.1*0.005 = 0.0005 moles

NH3(aq) + HCl (aq) -------------> NH4Cl(aq)

I 0.001 0.0005 0

C -0.0005 -0.0005 0.0005

E 0.0005 0 0.0005

Pkb = -logkb

= -log1.8*10^-5

= 4.75

POH = Pkb + log[NH4Cl]/[NH3]

= 4.75 + log0.0005/0.0005

= 4.75 + log1

= 4.75

PH = 14-POH

= 14-4.75 = 9.25 >>>>>answer

queen_honey_blossom answered 2 years ago

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Question

What is the pH of 10ml of 0.1M NH3, with 5ml of 0.1M of HCL. kb=1.8*10^-5

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