what is the resultig pH if 0.005 mol of NaOH is added to 100.0mL of buffer containing 0.010 mol acetic acid (CH3COOH,pKa = 4.74)and 0.010 mol sodium acetate (NaCH3COO)? (Assume the final volume is 100 mL)

So i worked it out, i just need to double check the pH. Let me know what you get, thank you.

Why don't you show what you have and how you obtained that and let US check you?

Oh, ok well i came out with a pH of 5.22. pH = pKa + log A/HA. I came up with .15M as A and .05 as HA. So 4.74+log(3)= 5.22.

Thanks. I agree.

To calculate the resulting pH, we need to consider the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the pKa of the weak acid and the ratio of its conjugate base to weak acid. The Henderson-Hasselbalch equation is given by:

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

where pH is the resulting pH, 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.

Given:
- Weak acid (acetic acid, CH3COOH) concentration: 0.010 mol
- Conjugate base (sodium acetate, NaCH3COO) concentration: 0.010 mol
- pKa of acetic acid (CH3COOH): 4.74

First, we need to calculate the concentrations of [A-] and [HA] after adding 0.005 mol of NaOH.

1. Calculate the moles of acetic acid and sodium acetate present initially:
Moles of acetic acid = concentration × volume = 0.010 mol
Moles of sodium acetate = concentration × volume = 0.010 mol

2. Calculate the moles of acetic acid and sodium acetate after adding NaOH:
Moles of acetic acid = initial moles - moles of NaOH added = 0.010 mol - 0.005 mol = 0.005 mol
Moles of sodium acetate = initial moles = 0.010 mol

3. Calculate the concentrations of [A-] and [HA]:
[A-] = moles of sodium acetate / volume = 0.005 mol / 0.100 L = 0.050 M
[HA] = moles of acetic acid / volume = 0.005 mol / 0.100 L = 0.050 M

4. Substitute the values into the Henderson-Hasselbalch equation:
pH = 4.74 + log(0.050/0.050)
pH = 4.74 + log(1)
pH = 4.74 + 0
pH = 4.74

Therefore, the resulting pH is 4.74.

Note: The final volume was assumed to be 100 mL, but it does not affect the calculations as the volume cancels out in the Henderson-Hasselbalch equation.