What is the pH of a 0.10 M solution of ( \text{NH}_4\text{F} )? The ( K_a ) of ( \text{NH}_4^+ ) is ( 5.6 \times 10^{-10} ) and the ( K_a ) of HF is ( 6.8 \times 10^{-4} ).
First, let's understand what is being asked.
We need to find the pH of a 0.10 M solution of ammonium fluoride, ( \text{NH}_4\text{F} ).
Given:
[ K_a(\text{NH}_4^+) = 5.6 \times 10^{-10} ]
Note that actually this is ( K_b(\text{NH}_3) ), but ( K_a(\text{NH}_4^+) = \frac{K_w}{K_b(\text{NH}_3)} ).
Also given:
[ K_a(\text{HF}) = 6.8 \times 10^{-4} ]
First, figure out what ions are present when ( \text{NH}_4\text{F} ) dissolves in water and how they affect the pH.
First, ( \text{NH}_4\text{F} ) dissociates:
[ \text{NH}_4\text{F} \rightarrow \text{NH}_4^+ + \text{F}^- ]
Now, both ( \text{NH}_4^+ ) and ( \text{F}^- ) are ions that can affect the pH.
- ( \text{NH}_4^+ ) is the conjugate acid of ( \text{NH}_3 ), so it can act as a weak acid:
[ \text{NH}_4^+ \rightleftharpoons \text{NH}_3 + \text{H}^+ ]
[ K_a(\text{NH}_4^+) = 5.6 \times 10^{-10} \text{ (given)} ]
- ( \text{F}^- ) is the conjugate base of ( \text{HF} ), which is a weak acid, so ( \text{F}^- ) acts as a weak base:
[ \text{F}^- + \text{H}_2\text{O} \rightleftharpoons \text{HF} + \text{OH}^- ]
[ K_b(\text{F}^-) = \frac{K_w}{K_a(\text{HF})} = \frac{1.0 \times 10^{-14}}{6.8 \times 10^{-4}} = 1.47 \times 10^{-11} ]
So both ( \text{NH}_4^+ ) and ( \text{F}^- ) can react with water, affecting the pH.
We have a salt of a weak acid and a weak base.
But we need to calculate the pH of the solution taking into account both.
In the presence of both ( \text{NH}_4^+ ) and ( \text{F}^- ), the solution's pH depends on the relative strengths of the conjugate acid ( \text{NH}_4^+ ) and conjugate base ( \text{F}^- ).
One standard way to approach this is to compare ( K_a(\text{NH}_4^+) ) and ( K_b(\text{F}^-) ).
But we may also write an expression for the equilibrium concentrations, set up an equilibrium calculation.
Alternatively, note that for salts of weak acids and weak bases, the pH depends on the ( K_a ) and ( K_b ) values.
One method is to use the formula:
[ \text{pH} = 7 + 0.5 (\text{p}K_a - \text{p}K_b) ]
But this formula works only when concentrations are equal.
Wait, perhaps more accurate to find ( K_b ) for ( \text{F}^- ) and compare it to ( K_a ) for ( \text{NH}_4^+ ).
Let's calculate ( K_b ) for ( \text{F}^- ).
Given ( K_a(\text{HF}) = 6.8 \times 10^{-4} ),
So ( K_b(\text{F}^-) = \frac{K_w}{K_a(\text{HF})} ).
Similarly, for ( \text{NH}_4^+ ), we can write ( K_b(\text{NH}_3) = \frac{K_w}{K_a(\text{NH}_4^+)} ).
But maybe not necessary.
Alternatively, for a salt of a weak acid and a weak base, the pH is given by:
[ \text{pH} = 7 + 0.5 \times \log \left( \frac{K_b}{K_a} \right) ]
But this formula assumes equimolar concentrations.
Wait, but in our case, the weak acid and weak base have the same concentration, because ( \text{NH}_4\text{F} ) dissociates into equal amounts of ( \text{NH}_4^+ ) and ( \text{F}^- ).
So let's compute:
[ \text{pH} = 7 + 0.5 \times \log \left( \frac{K_b}{K_a} \right) ]
Wait, the correct formula is:
[ \text{pH} = 7 + 0.5 \times \log \left( \frac{K_b \text{ (for base)}}{K_a \text{ (for acid)}} \right) ]
So first compute ( K_b ) for ( \text{F}^- ) and ( K_a ) for ( \text{NH}_4^+ ).
We have:
[ K_a(\text{NH}_4^+) = 5.6 \times 10^{-10} \text{ (given)} ]
We can compute:
[ K_b(\text{F}^-) = \frac{K_w}{K_a(\text{HF})} = \frac{1.0 \times 10^{-14}}{6.8 \times 10^{-4}} = 1.470588 \times 10^{-11} ]
Compute ( K_b(\text{F}^-) ):
[ K_b(\text{F}^-) = 1.470588 \times 10^{-11} ]
Now compute:
[ \text{pH} = 7 + 0.5 \times \log \left( \frac{K_b}{K_a} \right) ]
Plug in values:
[ \text{pH} = 7 + 0.5 \times \log \left( \frac{K_b(\text{F}^-)}{K_a(\text{NH}_4^+)} \right) ]
Plug in numbers:
[ \text{pH} = 7 + 0.5 \times \log \left( \frac{1.470588 \times 10^{-11}}{5.6 \times 10^{-10}} \right) ]
We can compute the ratio:
[ K_\text{ratio} = \frac{1.470588 \times 10^{-11}}{5.6 \times 10^{-10}} ]
[ K_\text{ratio} \approx \frac{1.470588}{5.6} \times \left( \frac{10^{-11}}{10^{-10}} \right) \approx (0.262605) \times (10^{-1}) = 0.0262605 ]
Now compute:
[ \log(K_\text{ratio}) = \log(2.62605 \times 10^{-2}) = -1.5800 ]
Then:
[ \text{pH} = 7 + 0.5 \times (-1.5800) = 7 - 0.79 = 6.21 ]
Therefore, the pH is approximately 6.21.
To determine the pH of a 0.10 M solution of ammonium fluoride (NH₄F), we'll need to consider the acid-base properties of its constituent ions when dissolved in water.
Step 1: Dissociation of NH₄F
[ \text{NH}_4\text{F} \rightarrow \text{NH}_4^+ + \text{F}^- ]
Both ( \text{NH}_4^+ ) and ( \text{F}^- ) can react with water, affecting the pH.
Step 2: Determine the Acid and Base Constants
- ( \text{NH}_4^+ ) is the conjugate acid of the weak base NH₃. It can donate a proton to water:
[ \text{NH}_4^+ \rightleftharpoons \text{NH}_3 + \text{H}^+ ]
The acid dissociation constant for ( \text{NH}_4^+ ) is given:
[ K_a(\text{NH}_4^+) = 5.6 \times 10^{-10} ]
- ( \text{F}^- ) is the conjugate base of the weak acid HF. It can accept a proton from water:
[ \text{F}^- + \text{H}_2\text{O} \rightleftharpoons \text{HF} + \text{OH}^- ]
First, calculate the base dissociation constant ( K_b ) for ( \text{F}^- ):
[ K_b(\text{F}^-) = \frac{K_w}{K_a(\text{HF})} = \frac{1.0 \times 10^{-14}}{6.8 \times 10^{-4}} = 1.47 \times 10^{-11} ]
Step 3: Use the pH Formula for a Salt of Weak Acid and Weak Base
For a solution containing equal concentrations of a weak acid and a weak base:
[ \text{pH} = 7 + \frac{1}{2} \log \left( \frac{K_b}{K_a} \right) ]
Plug in the values:
[ \frac{K_b}{K_a} = \frac{1.47 \times 10^{-11}}{5.6 \times 10^{-10}} = 0.02625 ]
Calculate the logarithm:
[ \log(0.02625) = -1.580 ]
Compute the pH:
[ \text{pH} = 7 + \frac{1}{2} (-1.580) = 7 - 0.790 = 6.21 ]
Answer:
[ 6.21 ]