What is the pH of a solution that consists of 0.25 M HC₂H₃O₂ and 0.35 M NaC₂H₃O₂?
K _a(HC₂H₃O₂) = 1.8×10^{- 5}

Question

What is the pH of a solution that consists of 0.25 M HC₂H₃O₂ and 0.35 M NaC₂H₃O₂? K _a(HC₂H₃O₂) = 1.8×10^{- 5}

Quizplus · Accepted Answer

This is a problem involving the dissociation of a weak acid, HCS1U1B12S1U1B0HS1U1B13S1U1B0OS1U1B12S1U1B0. The given equilibrium constant, K S1U1B1aS1U1B0(HCS1U1B12S1U1B0HS1U1B13S1U1B0OS1U1B12S1U1B0) = 1.8×10S1U1P1 - 5S1S1P0, can be used to calculate the pH of the solution.
To do this, we can use the equation for the dissociation of the weak acid:
HCS1U1B12S1U1B0HS1U1B13S1U1B0OS1U1B12S1U1B0 ⇌ H+ + CS1U1B12S1U1B0HS1U1B13S1U1B0OS1U1B12S1U1B0
Let x be the concentration of H+ ions.
Then we can set up the equilibrium expression:
K S1U1B1aS1U1B0 = [H+][CS1U1B12S1U1B0HS1U1B13S1U1B0OS1U1B12S1U1B0] / [HCS1U1B12S1U1B0HS1U1B13S1U1B0OS1U1B12S1U1B0]
Plugging in the given values:
1.8×10S1U1P1 - 5S1S1P0 = x[CS1U1B12S1U1B0HS1U1B13S1U1B0OS1U1B12S1U1B0] / [HCS1U1B12S1U1B0HS1U1B13S1U1B0OS1U1B12S1U1B0]
We also know that [HCS1U1B12S1U1B0HS1U1B13S1U1B0OS1U1B12S1U1B0] = 0.25 M and [NaCS1U1B12S1U1B0HS1U1B13S1U1B0OS1U1B12S1U1B0] = 0.35 M (since NaCS1U1B12S1U1B0HS1U1B13S1U1B0OS1U1B12S1U1B0 fully dissociates in solution).
So we can plug these into the expression and solve for x:
1.8×10S1U1P1 - 5S1S1P0 = x[0.35] / [0.25]
x = 2.52×10S1U1M1 - 5S1S1P0
Finally, we can take the negative logarithm of the H+ concentration to find the pH:
pH = -log[H+]
pH = -log(2.52×10S1U1M1 - 5S1S1P0)
pH = 4.89
Therefore, the answer is C, 4.89.

What Is the PH of a Solution That Consists of 0.25