A 1 liter solution contains 0.382 M hydrofluoric acid and 0.509 M sodium fluoride. Addition of...

A 1 liter solution contains 0.382 M hydrofluoric acid and 0.509 M sodium fluoride. Addition of 0.191 moles of sodium hydroxide will: (Assume that the volume does not change upon the addition of sodium hydroxide.) Raise the pH slightly, Lower the pH slightly, Raise the pH by several units, Lower the pH by several units, Not change the pH ,Exceed the buffer capacity

Expert Solution

no of moles of HF = molarity * volume in L

= 0.382*1 = 0.382moles

no of moles of NaF = molarity * volume in L

= 0.509*1 =- 0.509moles

PH = Pka + log[NaF]/[HF]

= 3.17+ log0.509/0.382

= 3.17+ 0.1246 = 3.2946

by the addition of 0.191 moles of NaOH

no of moles of HF = 0.382-0.191 = 0.191moles

no of moles of NaF = 0.509+0.191 = 0.7 moles

PH = Pka + log[NaF]/[HF]

= 3.17 + log0.7/0.191

= 3.17+ 0.564 = 3.734

Raise the pH slightly

queen_honey_blossom answered 2 years ago

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