Question

Calculate the expected ph of the following solutions. Please show all of your work and explain your steps, if you can.

Solution 1: 50mL 0.10M NH3 + 50mL 0.10M NH4NO3

Solution 2: 10mL solution 1 + 5mL H2O + 1mL 0.10M HCl

Solution 3: 10mL solution 1 + 6mL 0.10M HCl

Solution 4: 10mL solution 1 + 5mL H2O + 1mL 0.10M NaOH

Please write legibly so I can see what steps you took!

Answer 1

pK_a of NH₄⁺ = 9.25

Solution 1:

50 mL of 0.10 M of NH₃ = 50 x 0.10 = 5 mmol of NH₃

50 mL of 0.10 M of NH₄NO₃ = 50 x 0.10 = 5 mmol of NH₄NO₃

Total volume of the buffer solution = (50 + 50) = 100 mL

Concentration of NH₃ = 5/100 = 0.05 M of NH₃

Concentration of NH₄NO₃ = 5/100 = 0.05 M of NH₄NO₃

From Henderson-Hasselbalch equation

pH = pK_a + log[base]/[acid]

     = 9.25 + log 0.05/0.05

     = 9.25

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Solution 2:

1 mL of 0.10 M of HCl = 0.1 mmol of HCl

0.1 mmol of HCl will react with 0.1 mmol of NH₃ to form 0.1 mmol of NH₄⁺.

Now, 100 mL solution 1 contains 5 mmol of NH₃ and 5 mmol of NH₄NO₃

Therefore, 10 mL of solution 1 contains 0.5 mmol of NH₃ and 0.5 mmol of NH₄NO₃

After the addition of HCl, the solution 2 contains (0.5 - 0.1) = 0.4 mmol of NH₃ and (0.5 + 0.1) = 0.6 mmol of NH₄NO₃

Total volume of the solution 2 = 10 + 5 + 1 = 16 mL

Now, the concentration of NH₃ in solution 2 = 0.4/16 = 0.025 M

And the concentration of NH₄NO₃ in solution 2 = 0.6/16 = 0.0375 M

From Henderson-Hasselbalch equation

pH = pKa + log[base]/[acid]

     = 9.25 + log 0.025/0.0375

     = 9.25 - 0.18

     = 9.07

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Solution 3:

6 mL of 0.10 M of HCl = 0.6 mmol of HCl

Now, 10 mL of solution 1 contains 0.5 mmol of NH₃ and 0.5 mmol of NH₄NO₃

Of 0.6 mmol HCl, 0.5 mmol of HCl will react with 0.5 mmol of NH₃ to form 0.5 mmol of NH₄⁺

Remaining moles of HCl after neutralization = (0.6 - 0.5) = 0.1 mmol

Total volume of the solution 3 = 10 + 6 = 16 mL

Now, the concentration of remaining HCl in solution 3 = 0.1/16 = 0.00625 M

Thus,

pH = - log[H⁺]

     = - log 0.00625

    = 2.20

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Solution 4:

1 mL of 0.10 M of NaOH = 0.1 mmol of NaOH

0.1 mmol of NaOH will react with 0.1 mmol of NH₄⁺ to form 0.1 mmol of NH₃.

Now, 10 mL of solution 1 contains 0.5 mmol of NH₃ and 0.5 mmol of NH₄NO₃

After the addition of NaOH, the solution 4 contains (0.5 - 0.1) = 0.4 mmol of NH₄NO₃ and (0.5 + 0.1) = 0.6 mmol of NH₃

Total volume of the solution 4 = 10 + 5 + 1 = 16 mL

Now, the concentration of NH₄NO₃ in solution 4 = 0.4/16 = 0.025 M

And the concentration of NH₃ in solution 4 = 0.6/16 = 0.0375 M

From Henderson-Hasselbalch equation

pH = pK_a + log[base]/[acid]

     = 9.25 + log 0.0375/0.025

     = 9.25 + 0.18

     = 9.43