# Titration of Buffers: NH2OH and HCl

• Mitchtwitchita
In summary, the titration of a 0.350 M aqueous solution of the weak base NH2OH with 0.350 M HCl will result in 35 mL of HCl being added at the equivalence point. Before any acid is added, the pH of the solution is 9.68. After 10.00 mL of acid has been added, the pH of the solution is 6.22. At the equivalence point, the pH is 3.286. When 40.00 mL of acid has been added, the pH is 1.645.
Mitchtwitchita
You have 35.00 mL of a 0.350 M aqueous solution of the weak base NH2OH (Kb = 6.6 x 10^-9). This solution will be titrated with 0.350 M HCl.
(a)How many mL of acid must be added to reach the equivalence point?
(b)What is the pH of the solution before any acid is added?
(c)What is the pH of the solution after 10.00 mL of acid has been added?
(d)What is the pH of the solution at the equivalence point of the titration?
(e)What is the pH of the solution when 40.00 mL of acid has been added?

(a) Since the moles of acid and base are the same at the equivalence point, there will be 35 mL of HCl added.

(b)Kb = [NH3][OH-]/[NH2OH]
=6.6 x 10^-9 = x^2/.350
=4.8 x 10^-5
pOH = -log(4.8 x 10^-5)
=4.32
pH = 9.68

(c) moles NH2OH = 0.03500 L / 0.350 moles/L = 1.23 x 10^-2
moles HCl = 0.01000 L / 0.350 moles/L = 3.50 x 10^-3

NH2OH + HCl ---> NH3 + H20
Initial 1.23 x10^-2 3.50 x 10^-3 0
Change -3.50 x 10^-3 -3.50 x 10^-3 +3.50 x 10^-3
Equilibrium 8.80 x 10^-3 0 3.50 x 10^-3

NH2OH ---> NH3 + OH-

Kb = [NH3][OH-]/[NH2OH]
6.60 x 10^-9 = (3.50 x 10^-3)x/8.80 x 10^-3
[OH-] = 1.70 x 10^-8

pOH = -log[OH-]
=7.780
Therefore, pH = 6.220

(d) moles NH2OH = 1.23 x 10^-2
moles HCl = 1.23 x 10^-2

NH2OH + HCl ---> NH3 + H20
Initial 1.23 x 10^-2 1.23 x 10^-2 0
Change -1.23 x 10^-2 -1.23 x 10^-2 +1.23 x 10^-2
Equilibrium 0 0 1.23 x 10^-2

[NH2-] = 1.23 x 10^-2/0.0700 L = 0.176 M

NH3 + H2O ---> NH2- + H+
Initial 0.176 0 0
Change -x +x +x
Equilibrium 0.176 - x x x

Ka = Kw/Kb
=1.52 x 10^-6

Ka = [NH2-][H+]/[NH3]
1.52 x 10^-6 = x^2/0.176
[H+] = 5.17 x 10^-4

pH = -log[H+]
=3.286

(e)moles NH2OH = 1.23 x 10^-2
moles HCl = 0.04000 L / 0.350 moles/L = 1.40 x 10^-2

NH2OH + HCl ---> NH3 + H2O
Initial 1.23 x 10^-2 1.40 x 10^-2 0
Change -1.23 x 10^-2 -1.23 x 10^-2 +1.23 x 10^-2
Equilibrium 0 1.70 x 10^-3 1.23 x 10^-2

[H+] = 1.70 x 10^-3/0.07500 L = 2.27 x 10^-2 M

pH = -log[H+]
=1.645

Does this look right to anyone? If not could you please tell me where I'm going wrong? Also, do my significant digits look correct? I get confused on how they work with logs.

Lol...you go to Brock right?

Your calculations and significant digits look correct. However, it is always a good idea to double check your calculations and units to make sure they are accurate. Additionally, it is important to note that the pH at the equivalence point of a weak acid-strong base titration is not always 7. It can vary depending on the strength of the acid and base being titrated. Overall, your response is thorough and well-supported with calculations.

## 1. What is titration of buffers and why is it important?

Titration of buffers is a laboratory technique used to measure the concentration of an unknown solution by adding a known amount of another solution. It is important because it allows scientists to accurately determine the concentration of a solution and make precise calculations for experiments.

## 2. How do you prepare a buffer solution of NH2OH and HCl?

To prepare a buffer solution of NH2OH and HCl, you will need to mix the two solutions in a specific ratio based on the Henderson-Hasselbalch equation. For example, to make a buffer with a pH of 7, you would mix 10 mL of 0.1 M NH2OH with 90 mL of 0.1 M HCl.

## 3. What is the purpose of using a buffer in titration experiments?

The purpose of using a buffer in titration experiments is to maintain a stable pH level. Buffers are solutions that resist changes in pH when small amounts of acid or base are added. This allows for more accurate and precise results during titration.

## 4. How do you calculate the pH of a buffer solution of NH2OH and HCl?

The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where pKa is the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.

## 5. What are some common sources of error in a titration of NH2OH and HCl?

Some common sources of error in a titration of NH2OH and HCl include inaccurate measurement of solutions, improper mixing of solutions, and the presence of impurities in the solutions. It is important to carefully follow the experimental procedure and double-check measurements to minimize these errors.

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