a flask containing 50ml of 0.128M sodium acetate is titrated with 0.138M nitric acid. Calculate the initial ph of the acetate solution and calculate the ph after the addition of 24.9ml of nitric acid. pKa for acetate 4.757

a. I will call acetate Ac^-.

......Ac^- + HOH ==> HAc ==> OH^-
I...0.128.............0......0
C......-x.............x......x
E...0.128-x...........x......x

Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.128-x)
Solve for x = (OH^-) and convert to pH. I let mmol sand for millimols.
mmols Ac^ initially = mL x M = ?
mmols HNO3 added = mL x M = ?
The HNO3 neutralizes some of the OH^- and that forces the Ac^- to hydrolyze further.
mmols Ac^- remaining = mmols to start - mmols HNO3 added. This is the initial concn in the next step.

For the next step, H^+ from the HNO3 adds to the Ac^- so you have an acetate/acetic acid buffer.
Use the Henderson-Hasselbalch equation to solve for pH.
pH = pKa + log (base)/(acid)
........Ac^- + HNO3 ==> HAc
I.......?.......0........?
etc.

Post your work if you get stuck.

To calculate the initial pH of the acetate solution, we need to use the Henderson-Hasselbalch equation, which is given by:

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

Where:
- pH is the solution's pH
- pKa is the dissociation constant of the sodium acetate (4.757 in this case)
- [A-] is the concentration of the acetate ion (CH3COO-)
- [HA] is the concentration of the acetic acid (CH3COOH)

Given that we have a 0.128M sodium acetate solution, the initial concentration of CH3COO- is 0.128M. Since sodium acetate is the salt of acetic acid, we can assume that it completely dissociates, so [HA] is 0.

Plugging these values into the Henderson-Hasselbalch equation, we get:

pH = 4.757 + log(0.128/0)
pH = 4.757 + log(infinity)
pH = 4.757 + infinity
pH = infinity

Since the acetate solution does not contain any acetic acid initially, the pH is infinity or undefined.

Now, let's calculate the pH after the addition of 24.9ml of 0.138M nitric acid. To do this, we need to calculate the new concentration of acetic acid (CH3COOH) and acetate ion (CH3COO-) using the titration equation.

We know that the initial volume of the acetate solution is 50ml, and we add 24.9ml of nitric acid. So the total volume becomes 50ml + 24.9ml = 74.9ml.

Now, let's determine the moles of nitric acid used (n) using the equation:

n = C * V

Where:
- C is the concentration of nitric acid (0.138M)
- V is the volume of nitric acid used (24.9ml)

n = 0.138M * 0.0249L
n = 0.0034356 moles

Since sodium acetate and nitric acid react in a 1:1 ratio, the moles of acetic acid (CH3COOH) formed (x) will also be 0.0034356 moles.

The initial concentration of acetic acid (CH3COOH) is 0 because there was none at the beginning. After the reaction, there are 0.0034356 moles of acetic acid in a volume of 74.9ml. We need to convert this into a concentration in order to use the Henderson-Hasselbalch equation.

The new concentration of acetic acid ([HA]) is given by:

[HA] = moles of acetic acid (x) / total volume (V)

[HA] = 0.0034356 moles / 0.0749 L
[HA] = 0.04584 M

The concentration of the acetate ion ([A-]) is the initial concentration minus the amount that reacted.

[A-] = initial concentration - moles of acetic acid (x) / total volume (V)

[A-] = 0.128M - 0.0034356 moles / 0.0749 L
[A-] = 0.128M - 0.04584 M
[A-] = 0.08216 M

Now, we can use the Henderson-Hasselbalch equation again to calculate the new pH after the reaction:

pH = pKa + log([A-]/[HA])
pH = 4.757 + log(0.08216/0.04584)
pH = 4.757 + log(1.79)
pH = 4.757 + 0.2521
pH = 5.0091

Therefore, after the addition of 24.9ml of nitric acid, the pH of the acetate solution is approximately 5.0091.