Calculate the ph of a solution containing 0.15M HF (ka= 7.1x10^-4) and 0.26M NaF.

This is a common ion problem. The ionization of HF, a weak acid, is reduced (suppressed somewhat) by the addition of the F^- from the NaF. The F^- is the common ion.

......HF --> H^+ + F^-
I...0.15.....0.....0
C.....-x.....x.....x
E...0.15-x...x.....x

The NaF is 100% ionized; therefore, the F^- from that is 0.26M

Ka = (H^+)(F^-)/(HF)
Plug in the numbers and solve.
(HF) = 0.15-x
(H^+) = x
(F^-) = 0.26+x. The 0.26 is from the NaF and x from the HF.

Solve for x = (H^+) and convert to pH.
(F^-) =

To calculate the pH of this solution, we first need to determine the concentrations of the H+ and OH- ions in the solution. Here's a step-by-step calculation:

Step 1: Write the dissociation equation for HF in water:
HF ⇌ H+ + F-

Step 2: Write the equilibrium expression for the dissociation of HF:
Ka = [H+][F-] / [HF]

Step 3: Substitute the known values into the Ka expression:
7.1x10^-4 = [H+][0.26] / [0.15]

Step 4: Rearrange the equation to solve for [H+]:
[H+] = (7.1x10^-4 * 0.15) / 0.26

Step 5: Calculate the value of [H+] using the above equation:
[H+] ≈ 0.000408

Step 6: Since NaF is a strong electrolyte and completely dissociates in water to give Na+ and F- ions, we assume that [F-] from NaF is equal to 0.26M.

Step 7: To calculate the concentration of OH- ions, we can use the fact that Kw = [H+][OH-], where Kw is the autoionization constant of water (Kw = 1x10^-14 at 25°C). Since we know the value of [H+], we can rearrange the equation to solve for [OH-]:
[OH-] ≈ Kw / [H+]
[OH-] ≈ 1x10^-14 / 0.000408

Step 8: Calculate the value of [OH-] using the above equation:
[OH-] ≈ 2.46x10^-11

Step 9: Since water is neutral, the concentration of H+ ions is equal to the concentration of OH- ions in a neutral solution. Therefore, the pH of the solution can be calculated using the formula:
pH = -log[H+]

Step 10: Calculate the pH of the solution:
pH = -log(0.000408)
pH ≈ 3.39

Therefore, the pH of the solution containing 0.15M HF and 0.26M NaF is approximately 3.39.

To determine the pH of this solution, we need to calculate the concentration of H+ ions in the solution. Here's how you can do it step by step:

Step 1: Write down the equation representing the dissociation of HF in water according to its Ka value:
HF + H2O ⇌ H3O+ + F-

Step 2: Determine the initial concentrations of HF and NaF:
[H+] = [F-] = 0.26 M (since NaF is a strong electrolyte and dissociates fully)

Step 3: Set up an ICE table to track the changes in concentrations of HF, H3O+, and F-:

HF + H2O ⇌ H3O+ + F-
Initial: 0.15 M 0 M 0.26 M
Change: -x M +x M +x M
Equilibrium: 0.15 - x M x M 0.26 + x M

Step 4: Write the expression for the equilibrium constant, Ka:

Ka = [H3O+][F-] / [HF]

Step 5: Substitute the equilibrium concentrations into the Ka expression:

7.1x10^-4 = x * (0.26 + x) / (0.15 - x)

Step 6: Solve for x by assuming x is small compared to 0.15:

Since the concentration of HF is significantly greater than x, we can approximate (0.15 - x) as 0.15:

7.1x10^-4 = x * (0.26 + x) / 0.15

Step 7: Simplify and solve the quadratic equation:

0.001065 = (0.26x + x^2) / 0.15

0.001065 * 0.15 = 0.26x + x^2

0.00015975 = 0.26x + x^2

x^2 + 0.26x - 0.00015975 = 0

Step 8: Solve the quadratic equation using the quadratic formula or another appropriate method. The two possible values of x will be -0.010594 and 0.160594. Since we are dealing with concentrations, -0.010594 is not a valid solution.

Step 9: Calculate the concentration of H3O+ ions in the solution using x:

[H3O+] = x = 0.160594 M

Step 10: Finally, calculate the pH of the solution using the formula:

pH = -log[H3O+]

pH = -log(0.160594)

pH ≈ 0.795