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A 200.0 mL buffer solution is 0.360 M NH3 and 0.250 M NH4Br. What mass of...

A 200.0 mL buffer solution is 0.360 M NH3 and 0.250 M NH4Br. What mass of HCl does this buffer neutralize before pH falls below 8.80?

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Expert Solution

Volume of buffer = 200.0 ml

= 200 ml * 1.0 L/1000 ml

= 0.200 L

Number of moles = molarity * volume in L

Initial moles of NH3 = 0.360 x 0.200

= 0.072 mol

Initial moles of NH4+ = initial moles of NH4Br

= 0.250* 0.200

= 0.05 mol

We know that

pKb of NH3 = 1.8 x 10-5

Kw= 1.0 x 10 -14

pKa of NH4+ = Kw/pKb of NH3

= 1.0 x 10 -14/1.8 x 10-5

= 5.556 x 10-10

And pK a = - log Ka

= - log 5.556 x 10-10

= 9.25

NH3 + HCl => NH4+ + Cl-

Let us if x be the moles of HCl added

Moles of NH3 = 0.072 - x

Moles of NH4+ = 0.05 + x

Henderson-Hasselbalch equation:

pH = pKa + log([NH3]/[NH4+])

8.80 = 9.25 + log(moles of NH3/moles of NH4+)

8.80 = 9.25 + log ((0.072 – x )/( 0.05 + x))

- 0.45 = log ((0.072 – x )/( 0.05 + x))

((0.072 – x )/( 0.05 + x)) = 10^- 0.45

((0.072 – x )/( 0.05 + x))=0.355

0.072-x = 0.01775 + 0.355 x

0.05424= 1.355x

x = 0.0400 moles

Moles of HCl added = 0.400 mol

Mass of HCl added = moles x molar mass

= 0.0400 x 36.46 = 1.46 g

queen_honey_blossom answered 2 years ago
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