Strong base is dissolved in 665 mL of 0.400 M weak acid (Ka = 3.69 × 10-5) to make a buffer with a pH of 3.94. Assume that the volume remains constant when the base is added.

HA(aq) + OH-(aq) -> H2O(l) + A-(aq)
Calculate the pKa value of the acid and determine the number of moles of acid initially present.

I got 4.43 for pKa. And 0.266 for moles HA.

When the reaction is complete, what is the concentration ratio of conjugate base to acid?

I got 0.32 for the concentration.

How many moles of strong base were initially added?

This part I keep getting 0.128 but it keeps telling me it is wrong. Please help

4.43 is right for pKa and 0.266 (266 millimols) is correct for mols HA initially. The base/acid ratio is 0.3235 (I would carry it to at least one more place so let's make it 0.324). The last part I would do this way--I think it is easier than using the ratio.

........HA + OH^- ==> A^- + H2O
initial 266,0..0......0.......0
add ...........x..................
change....-x...-x.....x.......x
equil...266-x...0.....x.......x

Then substitute into the HH equation as
3.94 = 4.43 + log(x)/(266-x)
and solve for x. I get 65 millimoles base added. The I check these things to see if that gives me the right pH.
pH = 4.43 + log (65/266-65)
pH = 4.43 + log(65/201) = 3.94

thank great help! much appreciated

Well, calculating the number of moles of strong base initially added requires a different approach. Let me help you out with that.

We know the concentration of the weak acid solution is 0.400 M, and the volume is 665 mL (convert this to liters: 0.665 L).

To calculate the number of moles of acid initially present, we can use the equation:

moles = concentration x volume

Moles of acid = 0.400 M x 0.665 L = 0.266 moles

So, the number of moles of acid initially present is indeed 0.266 moles, as you correctly determined.

Now, to calculate the moles of strong base initially added, we need to consider the balanced equation:

HA(aq) + OH-(aq) -> H2O(l) + A-(aq)

From the equation, we can see that 1 mole of HA reacts with 1 mole of OH- to produce 1 mole of A-. Since the concentration of the base is not given, let's assume we have x moles of strong base initially.

Using the stoichiometry of the balanced equation, the moles of base reacting should be equal to the moles of acid initially present:

x moles of base = 0.266 moles of acid

Therefore, the number of moles of strong base initially added is 0.266 moles.

So, 0.266 moles of strong base were initially added, not 0.128 moles.

To calculate the number of moles of strong base initially added, we can use the Henderson-Hasselbalch equation:

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

Given that the pH is 3.94, we can rearrange the equation to solve for the concentration ratio [A-]/[HA]:

[A-]/[HA] = 10^(pH - pKa)

[A-]/[HA] = 10^(3.94 - 4.43)
[A-]/[HA] = 10^(-0.49)
[A-]/[HA] ≈ 0.316

The concentration ratio of the conjugate base to the acid is approximately 0.316.

Now, to determine the number of moles of strong base initially added, we need to multiply the initial volume (665 mL) by the concentration:

moles of strong base = volume * concentration
moles of strong base = 665 mL * 0.400 M

Since the volume is given in milliliters and the concentration is in moles per liter (M), we need to convert milliliters to liters:

1 L = 1000 mL
665 mL = 665/1000 L
665 mL = 0.665 L

moles of strong base = 0.665 L * 0.400 M
moles of strong base = 0.266 mol

Therefore, the correct number of moles of strong base initially added is 0.266 moles.

To find the number of moles of strong base initially added, we need to use the information given in the problem. In a buffer solution, the weak acid and its conjugate base are present in roughly equal concentrations.

We can start by calculating the concentration (in moles per liter) of the weak acid (HA) in the buffer solution using the given volume and molarity:

Concentration = Molarity x Volume = (0.400 mol/L) x (0.665 L) = 0.266 mol

Since the weak acid and its conjugate base are present in equal concentrations, the concentration of the conjugate base (A-) is also 0.266 mol.

According to the balanced chemical equation provided:

HA(aq) + OH-(aq) -> H2O(l) + A-(aq)

One mole of HA reacts with one mole of OH-. From the reaction, we know that the moles of strong base (OH-) added would be equal to the moles of the conjugate base (A-) formed.

Therefore, the number of moles of the initially added strong base is 0.266 mol.

Make sure to double-check your calculations to ensure accuracy. If you're still getting a different answer, check for any mistakes in the conversion or calculation steps.