Question

A 130.0 −mL buffer solution is 0.105 M in NH3 and 0.135 M in NH4Br. What mass of HCl can this buffer neutralize before the pH falls below 9.00? If the same volume of the buffer were 0.250 M in NH3 and 0.390 M in NH4Br, what mass of HCl could be handled before the pH falls below 9.00? Need answer in grams. PS the answer is not .092

Answer 1

PART A:

pKb of NH₃= 4.75

Initial number of moles of NH₃

= ([NH₃])(volume of buffer solution in L)

= (0.105)(0.130)

= 0.01365

Initial number of moles of NH₄

= ([NH₄⁺])(volume of buffer solution in L)

= (0.135)(0.130)

= 0.01755

Using Henderson-Hesselbalach equation;

pOH = pKb + log([NH₄⁺]/[NH₃])

        = 4.75 + log(0.135/0.105)

        = 4.75 + log(1.29)

        = 4.75 + 0.11

        = 4.86

pH = 14 – 4.86

     = 9.14

When the pH is 9.00,

pOH = 14 - 9.00 = 5.00

5.00 - 4.86 = log([NH₄⁺]/[NH₃])

0.14 = log([NH₄⁺]/[NH₃])

[NH₄⁺]/[NH₃]= 10^0.14= 1.38

Number of moles of NH₄⁺/number of moles of NH₃= 1.38

When HCl is added, the following reaction occurs:

NH₃(aq) + HCl(aq) → NH₄Cl(aq)

According to the equation, the ratio of the number of moles of NH₃ that reacts to that of HCl that reacts is 1:1. Let x be the number of moles of HCl added.

Number of moles of NH₃ remaining after the reaction = 0.01365– x

Number of moles of NH₄⁺ present after the reaction

= (initial number of moles of NH₄⁺)+(number of moles of HCl added)

= 0.01755+x

So, after addition of HCl

Number of moles of NH₄⁺/number of moles of NH₃= 1.38

(0.01755+x) / (0.01365– x)= 1.38

(0.01755+x) = 1.38 (0.01365– x)

0.01755+ x = 0.018837 – 1.38x

2.38x = 0.001287

x = 0.001287 / 2.38

x = 0.0005

Molar mass of HCl = 36.5

1 mole of HCl= 36.5g

0.0005 moles of HCl = 0.0005 x 36.5 = 0.018 g

Thus, if more than 0.018 g of HCl were added to this buffer solution, its pH would fall below 9.00.

PART B:

pKb of NH₃= 4.75

Initial number of moles of NH₃

= ([NH₃])(volume of buffer solution in L)

= (0.250)(0.130)

= 0.0325

Initial number of moles of NH₄

= ([NH₄⁺])(volume of buffer solution in L)

= (0.390)(0.130)

= 0.0507

Using Henderson-Hesselbalach equation;

pOH = pKb + log([NH₄⁺]/[NH₃])

        = 4.75 + log(0.390/0.250)

        = 4.75 + log(1.56)

        = 4.75 + 0.19

        = 4.94

pH = 14 – 4.94

     = 9.06

When the pH is 9.00,

pOH = 14 - 9.00 = 5.00

5.00 - 4.94 = log([NH₄⁺]/[NH₃])

0.06 = log([NH₄⁺]/[NH₃])

[NH₄⁺]/[NH₃]= 10^0.06= 1.15

Number of moles of NH₄⁺/number of moles of NH₃= 1.15

When HCl is added, the following reaction occurs:

NH₃(aq) + HCl(aq) → NH₄Cl(aq)

According to the equation, the ratio of the number of moles of NH₃ that reacts to that of HCl that reacts is 1:1. Let x be the number of moles of HCl added.

Number of moles of NH₃ remaining after the reaction = 0.0325– x

Number of moles of NH₄⁺ present after the reaction

= (initial number of moles of NH₄⁺)+(number of moles of HCl added)

= 0.0507+x

So, after addition of HCl

Number of moles of NH₄⁺/number of moles of NH₃= 1.15

(0.0507+x) / (0.0325– x)= 1.15

(0.0507+x) = 1.15 (0.0325– x)

0.0507+ x = 0.037375 – 1.15x

2.15x = -0.013325

x = 0.001287 / 2.15

x = 0.0006

Molar mass of HCl = 36.5

1 mole of HCl= 36.5g

0.0005 moles of HCl = 0.0006 x 36.5 = 0.0219 g

Thus, if more than 0.0219 g of HCl were added to this buffer solution, its pH would fall below 9.00.