Problem 16
Question
What effect does adding more NaF have on the pH and buffer capacity of an aqueous solution that is initially \(1.0 \mathrm{MHF}\) and \(0.50 \mathrm{M} \mathrm{NaF} ?\)
Step-by-Step Solution
Verified Answer
Answer: Adding more sodium fluoride (NaF) to the aqueous solution will increase the pH and buffer capacity of the solution.
1Step 1: Identify the components of the buffer system
In this exercise, HF will act as the weak acid and its conjugate base, F-, which is provided by NaF, will act as the weak base. This solution is a classic example of an acidic buffer system.
2Step 2: Write the ionization reaction for HF
Write the ionization equation for HF, which is a weak acid:
\(HF_(aq) + H_2O_(l) \rightleftharpoons H_3O^+_(aq) + F^-_(aq)\)
3Step 3: Use the Henderson-Hasselbalch equation to find the initial pH
Apply the Henderson-Hasselbalch equation for acidic buffer systems to determine the initial pH of the solution:
\(pH = pK_a + \log{\frac{[A^-]}{[HA]}}\)
First, we need to find the \(pK_a\) of HF; given that the \(K_a\) of HF is around \(7.2 \times 10^{-4}\), we can find the \(pK_a\) value as follows:
\(pK_a = -\log(K_a) = -\log(7.2 \times 10^{-4}) = 3.14\)
Now, plug the initial concentrations of HF and F- into the Henderson-Hasselbalch equation:
\(pH = 3.14 + \log{\frac{0.50}{1.0}} = 3.14 - 0.3 = 2.84\)
The initial pH of the solution is 2.84.
4Step 4: Find the new pH after adding more NaF
Determine the change in the concentration of F- when more NaF is added to the solution. Let's assume we add \(x\) moles of NaF, and the total volume of the solution remains constant:
New concentration of F- = \(0.50 + x\)
New concentration of HF = \(1.0 - x\)
Calculate the new pH by plugging the new concentrations into the Henderson-Hasselbalch equation:
\(new\,pH = 3.14 + \log{\frac{0.50 + x}{1.0 - x}}\)
As more NaF is added (\(x\) increases), the ratio of \(\frac{F^-}{HF}\) increases, and the new pH will be greater than the initial pH (2.84). Therefore, the pH of the solution increases as more NaF is added.
5Step 5: Analyse the effect on buffer capacity
Buffer capacity refers to the ability of the buffer system to resist changes in pH when an acid or base is added. In general, buffer capacity is at its maximum when the concentrations of the weak acid and its conjugate base are equal (\([A^-] = [HA]\)). As more NaF is added, the ratio of \(\frac{F^-}{HF}\) increases, which means the buffer system becomes more effective in neutralizing added acids, thereby increasing the buffer capacity.
In conclusion, adding more sodium fluoride (NaF) to the aqueous solution initially containing 1.0 M HF and 0.50 M NaF will increase the pH and buffer capacity of the solution.
Key Concepts
Henderson-Hasselbalch EquationWeak AcidConjugate BaseBuffer Capacity
Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is a formula used to estimate the pH of a buffer solution. It is particularly useful for calculating the pH of a solution containing a weak acid and its conjugate base, or a weak base and its conjugate acid. The equation is given by: \[ pH = pK_a + \log{\left( \frac{[A^-]}{[HA]} \right)} \] In this equation,
- \( pH \) is the measure of acidity or alkalinity of the solution.
- \( pK_a \) is the negative logarithm of the acid dissociation constant \( K_a \), which indicates the strength of the weak acid.
- \([A^-]\) represents the concentration of the conjugate base.
- \([HA]\) indicates the concentration of the weak acid.
Weak Acid
A weak acid, by definition, does not completely ionize in an aqueous solution. This means only a portion of the acid molecules donate protons (\(H^+\)) to water. For instance, hydrofluoric acid (HF) is a classic example of a weak acid. Its ionization in water is represented by the equation: \[ HF_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + F^-_{(aq)} \] In this equilibrium reaction, the double arrows indicate that both the forward (ionization of \(HF\)) and reverse reactions (recombination of ions to form \(HF\)) occur simultaneously. Thus, the concentration of hydrogen ions (and therefore the acidity) is relatively low compared to strong acids that fully dissociate. Knowing the \(K_a\) and hence \(pK_a\) (\(-\log(K_a)\)) helps in understanding how strong the weak acid is and its behavior in a buffer solution.
Conjugate Base
The conjugate base is what remains of an acid molecule after it loses a proton during the ionization process in water. For hydrofluoric acid (HF), the conjugate base is the fluoride ion \(F^-\). Understanding the role of the conjugate base is crucial in buffer solutions, as it helps maintain the solution's pH. Here is some more insight:
- The conjugate base works in tandem with its corresponding weak acid to neutralize additions of other acids or bases.
- In the presence of an added acid, the conjugate base can accept protons, mitigating significant changes in pH.
- Conversely, when a base is added, the weak acid component can donate protons to maintain the equilibrium.
Buffer Capacity
Buffer capacity is a measure of a buffer solution's ability to resist pH change upon the addition of an acidic or basic component. It describes how well a buffer can stabilize pH, a critical concept in maintaining the stability of chemical and biological systems. Here's what you should know:
- Buffer capacity is optimized when the ratio of the conjugate base \([A^-]\) and the weak acid \([HA]\) is close to 1.
- This balance enables the buffer to neutralize equal amounts of added acid or base.
- When additional NaF is added to a solution containing \(HF\) and \(F^-\), the concentration of the conjugate base increases, enhancing the buffer capacity.
- A higher buffer capacity means the system can handle more substantial fluctuations in pH without significant change to the overall acidity or basicity.
Other exercises in this chapter
Problem 14
Identify a suitable buffer system to maintain a pH of 3.0 in an aqueous solution.
View solution Problem 15
What does "buffer capacity" mean?
View solution Problem 17
Three buffers are separately prepared using equal concentrations of formic acid and sodium formate, hydrofluoric acid and sodium fluoride, and acetic acid and s
View solution Problem 18
Equal volumes of two buffers are prepared with equal concentrations of acid and conjugate base, but they use different weak acids with different p \(K_{\mathrm{
View solution