A buffer is prepared by adding 108 mL of 0.48 M NH3 to 143 mL of...

A buffer is prepared by adding 108 mL of 0.48 M NH3 to 143 mL of 0.30 M NH4NO3. What is the pH of the final solution?

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Expert Solution

no of moles of NH3 = molarity * volume in L

= 0.48*0.108 = 0.05184 moles

no of moles of NH4Cl = molarity * volume in L

= 0.3*0.143 = 0.0429moles

Pkb of NH3 = 4.75

POH = Pkb + log[NH4NO3]/[NH3]

= 4.75 + log0.0429/0.05184

= 4.75-0.0822 = 4.6678

PH = 14-POH

= 14-4.6678 = 9.3322 >>>>answer

queen_honey_blossom answered 1 year ago

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