buffer calculate the ph of the .1m propanoic acid Ka 1.35e-5
You can't get away with no caps in chemistry even though it makes it easier to type. The problem is that m means molality; M means molarity and the two are not the same. Do you have 0.1M or 0.1m solution? Second point I want to make. Neither 0.1m nor 0.1M are buffer solutions. Is this a buffer or not? If a buffer you don't have enough information to solve the problem EVEN if you distinguish between m and M.
To calculate the pH of a buffer solution, we need to consider the concentration of the acid (propanoic acid in this case) and its dissociation constant (Ka). A buffer solution is a mixture of a weak acid and its conjugate base that can resist changes in pH upon addition of small amounts of acid or base.
Let's start by writing down the dissociation reaction of propanoic acid (CH3CH2COOH):
CH3CH2COOH ⇌ CH3CH2COO- + H+
From the dissociation reaction, we can see that the concentration of the propanoic acid and its conjugate base (CH3CH2COO-) determine the pH of the buffer solution.
Given:
- Concentration of propanoic acid (CH3CH2COOH): 0.1 M
- Dissociation constant (Ka) of propanoic acid: 1.35 × 10^-5
To determine the pH, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the pKa (negative logarithm of Ka) and the ratio of the concentrations of the acid and its conjugate base:
pH = pKa + log([A-] / [HA])
First, let's calculate the pKa of propanoic acid:
pKa = -log(Ka) = -log(1.35 × 10^-5)
Next, substitute the values into the Henderson-Hasselbalch equation:
pH = pKa + log([A-] / [HA])
In a buffer solution, the concentration of the acid ([HA]) and its conjugate base ([A-]) are related to each other. In this case, initially, the concentration of the propanoic acid and its conjugate base are equal since propanoic acid fully dissociates in water.
Thus, [HA] = [A-] = 0.1 M.
Substituting the values into the equation:
pH = pKa + log(0.1 / 0.1)
Since the ratio of [A-] / [HA] is 1, log(1) equals 0, so the equation simplifies to:
pH = pKa + 0
Therefore, the pH of the buffer solution is equal to the pKa of the propanoic acid, which is obtained from the equation:
pH = -log(Ka)
Now, let's calculate the pH:
pH = -log(1.35 × 10^-5)
Using a calculator, we find that the pH is approximately 4.87.
So, the pH of the 0.1 M propanoic acid buffer solution with a Ka of 1.35 × 10^-5 is approximately 4.87.