A buffer solution is made as followed:

i)adding 13.50mL of 0.200mol/L sodium hydroxide to 50.00mL of 0.100mol/L propanoic acid.
ii)diluting the resulted buffer into a total volume of 100.00mL

Using IRE-C tables (if possible) calculate:
a)The pH of the buffer -before the dilution
b)The pH of the buffer- after the dilution
c) 1.50mL of 0.10mol/L hydrochloric acid solution is added to 25.00mL of the diluted buffer solution. Calculate the pH after the addition of the hydrochloric acid solution.

I did part a) but I am not getting the correct answer, which is given. I know that part b) is done the same way and when diluting the buffer into a total volume of 100.00mL it should not affect the pH. However, since I cannot get the right answer for part a) my answer for part b) is also incorrect.

Propanoic acid is CH3CH2COOH. Let's simplify that by calling it HPr.

HPr + NaOH ==> NaPr + H2O

I have no idea what an IRE-C table is.
moles NaOH = L x M = 0.01350 x 0.200 M = 0.027
moles HPr = L x M = 0.05000 x 0.100 = 0.005.
initial concns:
HPr = 0.005 moles/0.0635 = ??
NaOH = 0.0135/0.0635 = ??
pH = pKa + log [(base)/(acid)]
pH = substitute what you are using for pKa (that may be the problem) and values for concn base (NaPr) and acid (HPr). If this doesn't work out for you, repost but give the value you are using for Ka and pKa.

To calculate the pH of a buffer solution, you need to consider the equilibrium that occurs between the weak acid and its conjugate base. The Henderson-Hasselbalch equation is commonly used for this purpose, and it can be expressed as:

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

Where pKa is the negative logarithm of the acid dissociation constant (Ka), [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.

For part a), we need to calculate the pH of the buffer solution before the dilution. Let's break down the steps:

Step 1: Calculate the concentration of the weak acid (propanoic acid) and the conjugate base (sodium propanoate) after the mixing occurs.

Given:
Volume of sodium hydroxide (NaOH) = 13.50 mL
Concentration of NaOH = 0.200 mol/L
Volume of propanoic acid (HA) = 50.00 mL
Concentration of propanoic acid = 0.100 mol/L

First, convert the mL to L:
Volume of NaOH = 13.50 mL ÷ 1000 = 0.0135 L
Volume of propanoic acid = 50.00 mL ÷ 1000 = 0.0500 L

Next, calculate the number of moles of NaOH and propanoic acid:
Moles of NaOH = volume (L) × concentration (mol/L) = 0.0135 L × 0.200 mol/L = 0.0027 mol
Moles of propanoic acid = 0.0500 L × 0.100 mol/L = 0.0050 mol

Since NaOH and propanoic acid react in a 1:1 ratio, both will be fully consumed during the reaction. This means that the resultant buffer solution will have 0.0027 mol of sodium propanoate (A-) and 0.0027 mol of propanoic acid (HA) in a total volume of 63.50 mL.

Step 2: Calculate the concentration of the weak acid and the conjugate base in the final buffer solution.

Given:
Total volume of the buffer solution = 63.50 mL
Final volume after dilution = 100.00 mL

Convert the volumes to liters:
Total volume of the buffer solution = 63.50 mL ÷ 1000 = 0.0635 L
Final volume after dilution = 100.00 mL ÷ 1000 = 0.1000 L

To calculate the concentrations, divide the number of moles by the total volume:
Concentration of A- = 0.0027 mol ÷ 0.1000 L = 0.027 mol/L
Concentration of HA = 0.0027 mol ÷ 0.1000 L = 0.027 mol/L

Step 3: Calculate the pKa of propanoic acid.
Since pKa is not given, we need to find it. You can refer to an IRE-C table or use the Ka value to calculate pKa.

Propanoic acid (CH3CH2COOH) is a weak acid with a known Ka value of 1.34 × 10^(-5) at 25°C. To calculate pKa:

pKa = -log(Ka) = -log(1.34 × 10^(-5)) = 4.87

Step 4: Calculate the pH of the buffer before dilution using the Henderson-Hasselbalch equation.

pH = pKa + log([A-]/[HA])
pH = 4.87 + log([0.027]/[0.027]) = 4.87

Therefore, the pH of the buffer before dilution is 4.87.

Now, you can proceed to part b) and c) using the same approach. Remember that when diluting the buffer into a larger volume, the pH value should remain relatively unchanged.