Question

In: Chemistry

Calculate the ph of a solution that is 0.30 M in ammonia (NH3) and 0.20 M...

Calculate the ph of a solution that is 0.30 M in ammonia (NH3) and 0.20 M in ammonium chloride. Kb=1.76 x 10^-5

Then calculate the ph if you add 10 mL of 1.0 M HCl to 50 mL of your solution in the problem above.

Expert Solution

kb = 1.76*10^-5

Pkb = -logKb

= log1.76*10^-5

= 4.754

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

= 4.754 + log0.2/0.3

= 4.754-0.1760 = 4.578

PH = 14-POH

= 14-4.578 = 9.422 <<<<answer

no of moles of NH4Cl = molarity * volume in L

= 0.2*0.05 = 0.01 moles

no of moles of NH3 = molarity * volume in L

= 0.3*0.05 = 0.015 moles

no of moles of HCl = molarity *volume in L

= 1*0.01 = 0.01 moles

no of moles of NH4Cl after the addition of 0.01 moles of HCl = 0.01+0.01 =0.02 moles

no of moles of NH3 after the addition of 0.01 moles of HCl = 0.015-0.01 = 0.005 moles

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

= 4.754 + log0.02/0.005

= 4.754+0.602 = 5.356

PH = 14-POH

= 14-5.356 = 8.644 >>>answer

queen_honey_blossom answered 2 years ago

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