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Consider 1.0 \(\mathrm{L}\) of a solution that is 0.85 \(\mathrm{M} \mathrm{HOC}_{6} \mathrm{H}_{5}\) and 0.80 \(\mathrm{M} \mathrm{NaOC}_{6} \mathrm{H}_{5} .\left(K_{\mathrm{a}} \text { for } \mathrm{HOC}_{6} \mathrm{H}_{5}=1.6 \times 10^{-10} .\right)\) a. Calculate the pH of this solution. b. Calculate the pH after 0.10 mole of \(\mathrm{HCl}\) has been added to the original solution. Assume no volume change on addition of HCl. c. Calculate the pH after 0.20 mole of NaOH has been added to the original buffer solution. Assume no volume change on addition of NaOH.

Short Answer

Expert verified
The initial pH of the buffer solution is \(9.77\). After adding \(0.10\) mole of HCl, the pH becomes \(9.66\). Finally, after adding \(0.20\) mole of NaOH, the pH is \(10.02\).

Step by step solution

01

Identify the initial concentrations of the buffer solution

For this step, we already know that the solution is 0.85 M HOC鈧咹鈧 (acid) and 0.80 M NaOC鈧咹鈧 (base). The K鈧 for HOC鈧咹鈧 is 1.6 脳 10^(-10).
02

Calculate the initial pH of the buffer solution

To calculate the initial pH, we'll use the Henderson-Hasselbalch equation: pH = pK鈧 + log ([A鈦籡/[HA]) First, we need to find pK鈧. pK鈧 = -log(K鈧), so pK鈧 = -log(1.6 脳 10^(-10)) = 9.80. Now, we can find the pH: pH = 9.80 + log(0.80/0.85) = 9.80 + (-0.03) = \(9.77\).
03

Calculate the resulting concentrations after adding HCl

We have added 0.10 mole of HCl, which will react with NaOC鈧咹鈧 to form HOC鈧咹鈧. We can use the following stoichiometric relation: NaOC鈧咹鈧 + HCl 鈫 NaCl + HOC鈧咹鈧 Moles of base consumed = 0.10 mol New moles of HOC鈧咹鈧 = 0.85 + 0.10 = 0.95 New moles of NaOC鈧咹鈧 = 0.80 - 0.10 = 0.70 Now calculate the new concentrations from the new moles (assuming no volume change): [HOC鈧咹鈧匽 = 0.95 mol/L [NaOC鈧咹鈧匽 = 0.70 mol/L
04

Calculate the pH after adding HCl

Now, we can use the Henderson-Hasselbalch equation again for part b: pH = pK鈧 + log ([A鈦籡/[HA]) pH = 9.80 + log(0.70/0.95) = 9.80 - 0.14 = \(9.66\).
05

Calculate the resulting concentrations after adding NaOH

We have added 0.20 mole of NaOH, which will react with HOC鈧咹鈧 to form NaOC鈧咹鈧. HOC鈧咹鈧 + NaOH 鈫 NaOC鈧咹鈧 + H2O Moles of acid consumed = 0.20 mol New moles of HOC鈧咹鈧 = 0.85 - 0.20 = 0.65 New moles of NaOC鈧咹鈧 = 0.80 + 0.20 = 1.00 Now calculate the new concentrations from the new moles (assuming no volume change): [HOC鈧咹鈧匽 = 0.65 mol/L [NaOC鈧咹鈧匽 = 1.00 mol/L
06

Calculate the pH after adding NaOH

Now, we can use the Henderson-Hasselbalch equation again for part c: pH = pK鈧 + log ([A鈦籡/[HA]) pH = 9.80 + log(1.00/0.65) = 9.80 + 0.22 = \(10.02\). To summarize the results: a. The initial pH of the buffer solution is 9.77. b. The pH after adding 0.10 mole of HCl is 9.66. c. The pH after adding 0.20 mole of NaOH is 10.02.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is an essential tool in chemistry for understanding buffer solutions and their pH levels. This equation is particularly handy in biochemistry and medicine for preparing buffer solutions that maintain a stable pH within a certain range.

Here's the equation: \[ \text{pH} = \text{pK}_a + \log \left( \frac{[A^-]}{[HA]} \right) \]
  • \([A^-]\) represents the concentration of the conjugate base.
  • \([HA]\) represents the concentration of the weak acid.
  • pK鈧 is the negative logarithm of the dissociation constant (K鈧) and represents the strength of the acid.
To use this equation effectively, you must know the concentrations of the weak acid and its conjugate base, along with the K鈧 of the acid.
These inputs help you accurately calculate the pH of the buffer solution. The equation assumes that the ionization of the weak acid and its conjugate base are in a dynamic equilibrium, providing a resistance to pH changes, which is the hallmark of buffer solutions.
pH Calculation
Calculating the pH of a buffer solution is a straightforward process once you understand the relevant components.
For the initial solution in our exercise, given that the solution has 0.85 M HOC鈧咹鈧 (weak acid) and 0.80 M NaOC鈧咹鈧 (base), the Henderson-Hasselbalch equation becomes your go-to tool.

### Initial pH CalculationTo find the initial pH:1. Calculate pK鈧 using the K鈧 value provided: \[ \text{pK}_a = -\log(1.6 \times 10^{-10}) = 9.80 \]2. Use the equation: \[ \text{pH} = 9.80 + \log \left( \frac{0.80}{0.85} \right) \] Simplifying this gives a pH of 9.77.

As the buffer solution's concentrations change with the addition of an acid or base, repeat the process using the updated concentrations to find the new pH.
Remember, even small changes in the concentrations can lead to noticeable changes in the pH due to the logarithmic nature of the equation.
Acid-Base Reactions
Understanding acid-base reactions is crucial when working with buffer solutions. These reactions involve the transfer of protons (H鈦) between species, leading to changes in the composition of the solution.
In our exercise:
  • When HCl (a strong acid) is added, it reacts with the conjugate base NaOC鈧咹鈧, decreasing the base's concentration while increasing that of the acid HOC鈧咹鈧.
  • The reaction is: \( \text{NaOC}_6\text{H}_5 + \text{HCl} \to \text{NaCl} + \text{HOC}_6\text{H}_5 \)
  • When NaOH (a strong base) is added, it reacts with the weak acid HOC鈧咹鈧, decreasing the acid's concentration while increasing the base, NaOC鈧咹鈧.
  • The reaction is: \( \text{HOC}_6\text{H}_5 + \text{NaOH} \to \text{NaOC}_6\text{H}_5 + \text{H}_2\text{O} \)
These reactions illustrate the buffer's ability to resist drastic pH changes by transforming added strong acids or bases into their weaker counterparts, which dampens shifts in hydrogen ion concentration.
Chemical Equilibrium
In the context of buffer solutions, chemical equilibrium plays a vital role in maintaining consistent pH levels. A buffer solution consists of a weak acid and its conjugate base, which are in equilibrium.

### Dynamic Equilibrium This equilibrium allows the solution to maintain a stable pH: - When an acid is added, the equilibrium shifts towards the formation of more weak acid, using up the added H鈦 ions. - Conversely, when a base is added, the equilibrium favors the formation of more conjugate base, using up the hydroxide ions (OH鈦).
This balance is what gives buffers their ability to moderate pH changes.

### Le Ch芒telier's Principle This principle describes how a system at equilibrium responds to disturbances: - The system shifts in a direction that reduces the effect of the disturbance. For a buffer:
  • Adding acid will shift the equilibrium to the left, reducing the increase in H鈦.
  • Adding base will shift it to the right, neutralizing the added OH鈦.
The robust nature of buffer systems is a practical application of equilibrium concepts, ensuring that solutions can withstand additions of external acids or bases without significant changes in pH.

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Most popular questions from this chapter

In the titration of 50.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M}\) methylamine, CH_{3} \mathrm { NH } _ { 2 } \(\left(K_{\mathrm{b}}=4.4 \times 10^{-4}\right),\) with 0.50 \(\mathrm{M} \mathrm{HCl}\) , calculate the pH under the following conditions. a. after 50.0 \(\mathrm{mL}\) of 0.50\(M \mathrm{HCl}\) has been added b. at the stoichiometric point

Consider the blood buffer system discussed in the Exercise \(96 .\) Patients with severe diarrhea can have an excessive loss of sodium bicarbonate (sodium hydrogen carbonate). How would this affect the \(\mathrm{pH}\) of blood? Explain. What would be the treatment of such a condition?

Consider the titration of 100.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M} \mathrm{HONH}_{2}\) by 0.100 \(\mathrm{M} \mathrm{HCl} .\left(K_{\mathrm{b}} \text { for } \mathrm{HONH}_{2}=1.1 \times 10^{-8} .\right)\) a. Calculate the \(\mathrm{pH}\) after 0.0 \(\mathrm{mL}\) of \(\mathrm{HCl}\) has been added. b. Calculate the \(\mathrm{pH}\) after 25.0 \(\mathrm{mL}\) of \(\mathrm{HCl}\) has been added. c. Calculate the \(\mathrm{pH}\) after 70.0 \(\mathrm{mL}\) of HCl has been added. d. Calculate the \(\mathrm{pH}\) at the equivalence point. e. Calculate the pH after 300.0 \(\mathrm{mL}\) of HCl has been added. f. At what volume of HCl added does the pH = 6.04?

Some \(\mathrm{K}_{2} \mathrm{SO}_{3}\) and \(\mathrm{KHSO}_{3}\) are dissolved in 250.0 \(\mathrm{mL}\) of solution and the resulting \(\mathrm{pH}\) is \(7.25 .\) Which is greater in this buffer solution, the concentration of \(\mathrm{SO}_{3}^{2-}\) or the concentration of \(\mathrm{HSO}_{3}-7\) If \(\left[\mathrm{SO}_{3}^{2-}\right]=1.0 \mathrm{M}\) in this solution, calculate the concentration of \(\mathrm{HSO}_{3}\) .

A sample of a certain monoprotic weak acid was dissolved in water and titrated with 0.125\(M \mathrm{NaOH}\) , requiring 16.00 \(\mathrm{mL}\) to reach the equivalence point. During the titration, the pH after adding 2.00 \(\mathrm{mL}\) . NaOH was 6.912 . Calculate \(K_{\mathrm{a}}\) for the weak acid.

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