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A 130.0 −mL buffer solution is 0.100 M in NH3 and 0.135 M in NH4Br. a.Part...

A 130.0 −mL buffer solution is 0.100 M in NH3 and 0.135 M in NH4Br. a.Part A What mass of HCl can this buffer neutralize before the pH falls below 9.00? b.Part B If the same volume of the buffer were 0.255 M in NH3 and 0.400 M in NH4Br, what mass of HCl could be handled before the pH falls below 9.00?

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Expert Solution

Part A)

pH = 9.00

pKa = 9.26

moles of NH3 = 0.130 x 0.100 = 0.0130

moles of NH4Br = 0.130 x 0.135 = 0.01755

NH3 (aq)   +    H+ (aq)   <---------------> NH4+ (aq)

0.0130              x                                    0.01755

0.0130 - x         0                                    0.01755 + x

pH = pKa + log [base / acid]

9.0 = 9.26 + log [0.0130 - x / 0.01755 + x ]

[0.0130 - x / 0.01755 + x ] = 0.55

0.0130 - x = 0.00965 + 0.55 x

x = 0.00216

moles of HCl = 0.00216

mass of HCl = 0.00216x 36.45

mass of HCl = 0.0788 g

B)

moles of NH3 = 0.130 x 0.255 = 0.03315

moles of NH4Br = 0.0520

NH3 (aq)   +    H+ (aq)   <---------------> NH4+ (aq)

0.03315            x                                    0.0520

0.03315 - x         0                                    0.0520 + x

pH = pKa + log [base / acid]

9.0 = 9.26 + log ( 0.03315 - x / 0.0520 + x )

( 0.03315 - x / 0.0520 + x ) = 0.55

0.03315 - x = 0.0286 + 0.55x

x = 0.00294

moles of HCl = 0.00294

mass of HCl = 0.107 g

queen_honey_blossom answered 2 years ago
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