A solution is prepared to be 0.10 M acetic acid, HC2H3O2, and 0.20 M sodium acetate, NaC2H3O2. What is the pH of this buffer? Ka for acetic acid is 1.7 x 10^-5. What is the pH after 9.5 mL of 0.10 M hydrochloric acid is added?
To find the pH of the buffer solution, we need to use the Henderson-Hasselbalch equation, which is:
pH = pKa + log([A-]/[HA])
Where:
- pH is the negative logarithm of the hydrogen ion concentration.
- pKa is the negative logarithm of the acid dissociation constant, Ka.
- [A-] is the concentration of the conjugate base (sodium acetate).
- [HA] is the concentration of the acid (acetic acid).
Given:
- Concetration of acetic acid, [HA] = 0.10 M
- Concentration of sodium acetate, [A-] = 0.20 M
- Ka for acetic acid = 1.7 x 10^-5
First, calculate pKa using the given Ka value:
pKa = -log(Ka) = -log(1.7 x 10^-5) = 4.77
Next, substitute the values into the Henderson-Hasselbalch equation:
pH = 4.77 + log(0.20/0.10)
= 4.77 + log(2)
= 4.77 + 0.301
= 5.071
Therefore, the pH of the buffer solution is approximately 5.071.
To find the pH after adding 9.5 mL of 0.10 M hydrochloric acid (HCl) to the buffer, we need to consider the changes in concentration of the acid and its conjugate base.
First, we need to convert the volume of HCl from mL to L:
Volume of HCl = 9.5 mL = 9.5/1000 L = 0.0095 L
The initial concentration of acetic acid, [HA] = 0.10 M, remains the same.
The concentration of sodium acetate, [A-], will change due to the reaction with HCl. For every 1 mole of HCl added, 1 mole of sodium acetate will be converted to acetic acid.
Next, we need to determine the number of moles of HCl added:
Number of moles of HCl = concentration of HCl x volume of HCl
= 0.10 M x 0.0095 L
= 0.00095 moles
Since the reaction between HCl and sodium acetate is 1:1, the concentration of sodium acetate ([A-]) will decrease by the same amount as the HCl concentration.
Now, calculate the new concentration of sodium acetate:
New [A-] concentration = initial [A-] - moles of HCl added / total volume of the solution
= 0.20 M - 0.00095 moles / (0.0095 L + 0.0095 L)
= 0.20 M - 0.00095 moles / 0.019 L
≈ 0.197 M
Then, substitute the new concentrations into the Henderson-Hasselbalch equation to find the new pH:
pH = pKa + log([A-]/[HA])
= 4.77 + log(0.197/0.10)
Using a calculator, you can find the value of log(0.197/0.10), which is approximately 0.0902.
Therefore, the new pH of the solution after adding 9.5 mL of 0.10 M HCl is:
pH = 4.77 + 0.0902
= 4.8602
Thus, the pH after adding 9.5 mL of 0.10 M HCl is approximately 4.8602.