Stuck on chemistry problem

Question

What is the pH and pOH of a solution that was made by adding 245 mL of water to 500 mL of 4 * 10 ^ - 4 M HF solution?

PH?round to 1 decimal place

pOH 1 decimal place

Answer 1

For this question you'll need the K_a of HF, which is reported online as 7.2x10^-4

We need to start by figuring out the concentration of our HF once we dilute it by adding water. This can be done by converting the original concentration and volume into moles of HF and then dividing by the new solution volume:

0.500 L(4.0x10^-4 mol/L) = 2.0x10^-4 mol HF (note that we converted 500 mL to 0.5 L)

2.0x10^-4 mol HF÷(0.245 L + 0.500 L) = 2.685x10^-4 mol/L HF (solution volume = 245 mL + 500 mL)

This concentration can be used to form an ICE table. The ICE table will ultimately give the pH and pOH of the solution:

HF + H₂O ------> H₃O⁺ + F^-

I 2.685x10^-4 -- 0 0

C -X -- +X +X

E 2.685x10^-4-X -- X X

This gives rise to a K_a expression that can be used to determine the concentration of H₃O⁺:

K_a = [Products]÷[Reactants]

K_a = (X)(X)÷(2.685x10^-4-X) = X²÷(2.685x10^-4-X)

We looked up the K_a for HF earlier so that we could plug it into this equation to solve for X:

7.2x10^-4 = X²÷(2.685x10^-4-X) (we want to set one side equal to zero to get a polynomial)

X²+7.2x10^-4X-1.933x10^-7 = 0 (we now want to use the quadratic formula to find where X = 0)

X = -9.282x10^-4 or 2.082x10^-4

Here we two values for X, however, only one is the correct answer. If we were to use the negative value for X and plug that back into our ICE table, we would get a negative H₃O⁺ concentration. Because we can't have negative concentrations, this answer is discarded. Instead, we use X = 2.082x10^-4 and plug this value back into our ICE table:

HF + H₂O ------> H₃O⁺ + F^-

E 2.685x10^-4-(2.082x10^-4) -- 2.082x10^-4 2.082x10^-4

Here, we can see that at equilibrium our H₃O⁺ concentration will be 2.082x10^-4 M. To find pH, we take the -log() of our H₃O⁺ concentration as p refers to the mathematical operation -log()

-log(2.082x10^-4) = 3.7 = pH

To find pOH, we can use the equation: pH+pOH = 14

14-3.7 = 10.3 = pOH