Calculate the pH if 0.03 mol HCl is added to 0.500 L of a buffer solution...

Calculate the pH if 0.03 mol HCl is added to 0.500 L of a buffer solution that is 0.24 M NH3 and 0.20 M NH4Cl?

For NH3 Kb= 1.8 x 10-5

Solutions

Expert Solution

no of moles of NH3 = molarity * volume in L

= 0.24*0.5 = 0.12 moles

no of moles of NH4Cl = molarity * volume in L

= 0.2*0.5 = 0.1 moles

PKb = -logKb

= -log1.8*10^-5

= 4.75

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

= 4.75 + log0.1/0.12

= 4.75-0.0791

= 4.6709

By the addition of 0.03 moles of HCl

no of moles of NH3 after addition of 0.03 moles of HCl = 0.12-0.03 = 0.09 moles

no of moles of NH4Cl after addition of 0.03 moles of HCl = 0.1+0.03 = 0.13 moles

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

= 4.75 + log0.13/0.09

= 4.75 + 0.1597 = 4.9097

PH = 14-POH

= 14-4.9097 = 9.0903

queen_honey_blossom answered 1 year ago

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