A 30 mL sample of .165 M propanoic acid is titrated with .300M KOH. Calculate the pH at each volume of added base: 0 mL, 5 mL, 10, equivalence point, one-half equivalence point, 20 mL, 25 mL. Use calculations to make a sketch of the titration curve.

To calculate the pH at each volume of added base in the titration of propanoic acid with KOH, you need to understand the stoichiometry of the reaction and the properties of the propanoic acid. Let's break down the process step by step:

1. First, you need to determine the balanced chemical equation for the reaction between propanoic acid (HA) and KOH (base):

HA + KOH -> K(A) + H2O

In this reaction, one mole of propanoic acid reacts with one mole of KOH, resulting in the formation of one mole of the potassium salt of propanoic acid (K(A)) and one mole of water (H2O).

2. Next, you need to determine the amount of propanoic acid and KOH present in the initial 30 mL sample.

Amount of propanoic acid = Molarity of propanoic acid (M) * Volume of propanoic acid (L)
Amount of KOH = Molarity of KOH (M) * Volume of KOH (L)

Given:
- Molarity of propanoic acid (HA) = 0.165 M
- Volume of propanoic acid (HA) = 30 mL = 0.03 L
- Molarity of KOH = 0.300 M

Amount of propanoic acid = 0.165 M * 0.03 L = 0.00495 moles
Amount of KOH = 0.300 M * Volume of KOH (L)

3. Now, let's calculate the pH at each volume of added base:

a) 0 mL of added base (before titration):
At this point, no KOH has been added, so the concentration of propanoic acid remains unchanged. The pH can be calculated by using the dissociation constant (pKa) of propanoic acid, which is 4.87.

pH = pKa + log([A-]/[HA])
Since no KOH has been added, [A-] = 0 and [HA] = 0.00495 M (amount of propanoic acid).

pH = 4.87 + log(0/[0.00495]) = 4.87

b) 5 mL of added base:
At this point, a fraction of the propanoic acid has reacted with a fraction of the KOH. To determine the amount of propanoic acid and KOH remaining:

Amount of propanoic acid remaining = Initial amount - Amount reacted
Amount of KOH reacted = Molarity of KOH * Volume of KOH reacted (L)

The calculation gets complicated and iterative at this point as the reaction progresses, so it is often helpful to use a tool like a spreadsheet or a software program like Excel to do the calculations.

c) Repeat the above calculation for 10 mL, the equivalence point, one-half equivalence point, 20 mL, and 25 mL of added base.

Note: The equivalence point is the point at which the number of moles of acid is stoichiometrically equal to the number of moles of base added. In other words, it is the point at which all the propanoic acid has reacted completely with KOH. The one-half equivalence point is the point at which half of the moles of propanoic acid have reacted with KOH.

4. Once you have calculated the pH at each volume of added base, you can plot a titration curve. On the x-axis, you will have the volume of added base, and on the y-axis, you will have the pH. Start with a pH of 4.87 at 0 mL, and as you add increments of base, record the pH values to create a curve.

Note that during titration, the pH usually changes sharply around the equivalence point. Before the equivalence point, the pH will gradually increase, and after the equivalence point, it will gradually decrease.

Remember, this is a basic outline of the calculations involved in determining the pH at each volume of added base in the titration of propanoic acid with KOH. Using specialized software or spreadsheets can make the calculations more manageable.

eq pt.

mL acid x M acid = mL KOH x M KOH
Solve for mL KOH.

0 mL:
.........HPr ==> H^+ + Pr^-
I......0.165.....0.....0
C.........-x......x....x
E.......0.165-x...x....x

Ka = (H^+)(Pr^-)/(HPr).
Substitute from the E line and solve for x = (H^+), then convert to pH.

For additions up to but not including eq pt use the Henderson-Hasselbalch equation.

For the eq pt the pH is determined by the hydrolysis of the salt. (salt) = (Pr^-) = (0.165 x 30 mL)/(total mL) where total mL = 30 + mL to eq pt. I will call this 0.1 M but you need to do it more accurately
........Pr^- + HOH ==> HPr + OH^-
I......0.1..............0.....0
C........-x.............x.....x
E......0.1-x............x.....x

Kb for Pr^- = (Kw/Ka for acid) = (x)(x)/(0.1-x). Solve for x = (OH^-) and convert to pH.

For all points past the eq pt it will be excess OH^-.
mols base = M x L
mols acid to eq pt = ?
excess OH^- = difference.
(OH^-) = mols OH excess/total L.
Then convert to pH.

2.8?