The pH of 0.40 M HF(aq) is 1.93. Calculate the change in pH when 0.301 g of sodium fluoride is added to 53.7 mL of the solution. Ignore any change in volume.

Well, let's see what we're dealing with here. First, we have a solution of HF(aq) with a pH of 1.93. Adding sodium fluoride (NaF) will result in the formation of more fluoride ions (F⁻). Now, you might think fluoride is such a tiny little thing, what difference could it possibly make? But let me tell you, the little guys pack a punch!

Since we're ignoring any change in volume, we only need to focus on the change in moles of fluoride ions. To do that, we need to convert grams of NaF to moles. So, let's do some quick calculation.

First, we need to find the molar mass of NaF. Na has a molar mass of 22.99 g/mol, and F has a molar mass of 18.998 g/mol.

So the molar mass of NaF = 22.99 + 18.998 = 41.988 g/mol.

Now, we can calculate the moles of NaF in 0.301 g: moles = mass/molar mass = 0.301/41.988 = 0.00718 mol.

Since NaF dissociates fully in water, this means we'll have 0.00718 mol of additional fluoride ions (F⁻) in our solution.

Now, here comes the pH action! HF is a weak acid that reacts with water to form H₃O⁺ and F⁻ ions. The initial concentration of HF is 0.40 M, and we've determined that we have an additional 0.00718 mol of F⁻ ions.

Since HF and F⁻ ions are in a 1:1 ratio, this means we're doubling the total concentration of F⁻ ions. So, the concentration of F⁻ ions becomes 0.40 + 0.40 = 0.80 M.

To find the new pH, we need to calculate the new H₃O⁺ concentration using the new concentration of F⁻ ions. However, since HF is a weak acid, the dissociation is not complete. So, we need to consider the equilibrium expression for HF:

HF ⇌ H⁺ + F⁻

The equilibrium constant (Ka) for this reaction is 6.8 x 10⁻⁴, which means that only a small fraction of HF molecules will dissociate into H⁺ and F⁻ ions.

The equilibrium expression for Ka is as follows:

Ka = [H⁺][F⁻]/[HF]

However, since we're ignoring any change in volume, the volume of the solution remains constant. This means that the initial concentration of HF remains the same, while the concentrations of H⁺ and F⁻ ions change.

Now, we can substitute the known values into the equilibrium expression:

6.8 x 10⁻⁴ = [H⁺][F⁻]/0.40

Since HF and F⁻ ions are in a 1:1 ratio, the concentrations of H⁺ ions and F⁻ ions are equal.

So, [H⁺] = [F⁻].

We can substitute this into the equation:

6.8 x 10⁻⁴ = [H⁺][H⁺]/0.40

6.8 x 10⁻⁴ = [H⁺]²/0.40

After solving this equation, we find that [H⁺] ≈ 1.0077 x 10⁻³ M.

Now, we can calculate the pH using the equation: pH = -log[H⁺].

pH = -log(1.0077 x 10⁻³) ≈ 2.996.

So, the change in pH is approximately 1.93 - 2.996 = -1.066.

Therefore, the change in pH, when 0.301 g of sodium fluoride is added to 53.7 mL of the solution, is approximately -1.066.

To calculate the change in pH when sodium fluoride (NaF) is added to the HF solution, we need to consider the dissociation reaction of HF and the reaction of NaF with HF. First, let's determine the moles of HF present in the solution.

Molar mass of HF = 20.01 g/mol
Mass of HF = concentration (M) × volume (L) = 0.4 M × 0.0537 L = 0.02148 mol

Since HF is a weak acid, it partially dissociates in water to form H+ and F- ions.

HF(aq) ⟶ H+(aq) + F-(aq)

Since the concentration of H+ is equal to the concentration of HF, we can use the Henderson-Hasselbalch equation to calculate the pH:

pH = pKa + log ([A-] / [HA])

The pKa value for HF is given by:
pKa = -log(Ka)

Ka is the acid dissociation constant, and for HF, it is 7.2 × 10^-4.

pKa = -log(7.2 × 10^-4) = 3.14

Let's calculate [A-] and [HA]:
[HA] = 0.02148 mol
[A-] = 0.02148 mol

Now we can calculate the initial pH:

pH = 3.14 + log (0.02148 / 0.02148)
pH = 3.14 + log(1)
pH = 3.14

Next, let's determine the moles of NaF added to the system. The molar mass of NaF is 41.99 g/mol.

Moles of NaF = mass / molar mass = 0.301 g / 41.99 g/mol = 0.00717 mol

Since NaF reacts with HF to form Na+ and F- ions, the reaction can be written as follows:

NaF(aq) + HF(aq) ⟶ Na+(aq) + F-(aq) + HF(aq)

From the balanced reaction, we can see that the moles of F- ions produced are the same as the moles of NaF added.
Thus, [A-] is increased by 0.00717 mol.

To calculate the change in pH, we need to recalculate using the new [A-] value.

pH = 3.14 + log ([A-] / [HA]) = 3.14 + log ([0.02148 + 0.00717] / 0.02148)

pH = 3.14 + log (0.02865 / 0.02148)
pH = 3.14 + log (1.3356)

Using a calculator, we can find that log (1.3356) is approximately 0.126.

pH = 3.14 + 0.126
pH = 3.266

Therefore, the change in pH when 0.301 g of sodium fluoride is added to the solution is approximately 0.126 units. The new pH is 3.266.

To calculate the change in pH when sodium fluoride is added to the solution, we need to consider the reaction between HF and NaF. In this case, HF acts as an acid, and NaF acts as its conjugate base.

The balanced equation for the reaction between HF and NaF is:

HF(aq) + NaF(aq) -> NaHF2(aq)

When NaF is added to the solution, it reacts with HF to form NaHF2. This reaction consumes some of the HF, resulting in a decrease in the concentration of HF, which in turn affects the pH of the solution.

First, let's calculate the initial concentration of HF using the given molarity and volume.

Initial concentration of HF:
Molarity = 0.40 M
Volume = 53.7 mL = 0.0537 L

Initial moles of HF = initial concentration * volume:
moles of HF = 0.40 M * 0.0537 L = 0.02148 mol

Since HF is a weak acid, it partially dissociates in water, which will affect the pH of the solution. We need to calculate the initial concentration of H+ ions in the solution:

HF(aq) ⇌ H+(aq) + F-(aq)

Ka (acid dissociation constant) for HF is 3.5 x 10^-4.

Using the equation for Ka, we can calculate the initial concentration of H+ ions:

Ka = [H+][F-] / [HF]

[H+] = (Ka * [HF]) / [F-]
[H+] = (3.5 x 10^-4 * 0.40 M) / 0.40 M
[H+] = 3.5 x 10^-4 M

The pH of the solution is given as 1.93. The relationship between pH and [H+] is given by the equation:

pH = -log[H+]

Taking the antilog of both sides, we have:

[H+] = 10^(-pH)
[H+] = 10^(-1.93)
[H+] = 7.37 x 10^(-2) M

Now, let's calculate the final concentration of HF and the concentration of H+ ions after adding NaF.

To do this, we need to calculate the moles of NaF introduced into the solution using the given mass and molar mass:

Mass of NaF = 0.301 g
Molar mass of NaF = 41.99 g/mol

Moles of NaF = mass / molar mass:
moles of NaF = 0.301 g / 41.99 g/mol = 0.00718 mol

Since NaF reacts with HF in a 1:1 stoichiometric ratio, the moles of HF consumed are equal to the moles of NaF added.

Now, let's calculate the final concentration of HF:

Final moles of HF = initial moles of HF - moles of NaF:
moles of HF = 0.02148 mol - 0.00718 mol = 0.01430 mol

Final concentration of HF:
Final concentration = final moles of HF / volume
Final concentration = 0.01430 mol / 0.0537 L = 0.266 M

Finally, let's calculate the final concentration of H+ ions and the change in pH:

[H+] = (Ka * [HF]) / [F-]
[H+] = (3.5 x 10^-4 * 0.266 M) / 0.266 M
[H+] = 3.5 x 10^-4 M

New pH = -log[H+]
New pH = -log(3.5 x 10^-4)
New pH = 3.46

Change in pH = New pH - Initial pH
Change in pH = 3.46 - 1.93
Change in pH = 1.53

Therefore, the change in pH when 0.301 g of sodium fluoride is added to 53.7 mL of the 0.40 M HF solution is approximately 1.53.

Use the Henderson-Hasselbalch equation.

pH = pKa + log(base)/(acid)

pKa you know or can look up.
base = mols NaF = grams/molar mass
acid = mols HF = M x L
Solve for pH and take the difference between the new pH and 1.93 to find the change.
Post your work if you get stuck.