Use the Henderson-Hasselbalch equation to calculate the pH of a buffer prepared by adding 10mL of acetic acid to 20mL of 0.10M sodium acetate (Ka= 1.75 x 10^5 for acetic acid).

Please showing working out, as am confused as to how to approach this question. Thanks in advance for your help.

pH = pKa + log [(base)/(acid)]

pKa = -log Ka
base = acetate
acid = acetic acid.

To calculate the pH of the buffer solution using the Henderson-Hasselbalch equation, we need to know the concentrations of the acid and its conjugate base, as well as the dissociation constant (Ka) for the acid.

Given:
- Volume of acetic acid (acid) = 10 mL
- Volume of sodium acetate (conjugate base) = 20 mL
- Concentration of sodium acetate = 0.10 M
- Dissociation constant (Ka) for acetic acid = 1.75 x 10^(-5)

First, we need to calculate the moles of acid and conjugate base:
Moles of acid = (volume of acid in liters) x (concentration of acid)
Moles of acid = (10 mL / 1000 mL/L) x 1.75 x 10^(-5) M = 1.75 x 10^(-7) moles

Moles of conjugate base = (volume of conjugate base in liters) x (concentration of conjugate base)
Moles of conjugate base = (20 mL / 1000 mL/L) x 0.10 M = 0.002 moles

Next, we substitute these values into the Henderson-Hasselbalch equation:

pH = pKa + log10 ([conjugate base] / [acid])

pKa = -log10(Ka) = -log10(1.75 x 10^(-5)) ≈ 4.76

pH = 4.76 + log10(0.002 / 1.75 x 10^(-7))

To simplify the calculation, we can convert 0.002 to scientific notation:
pH = 4.76 + log10(2 x 10^(-3) / 1.75 x 10^(-7))

Simplifying the ratio inside the log term:
pH = 4.76 + log10(2 / 1.75) + log10(10^(-3) / 10^(-7))

pH = 4.76 + log10(1.143) + log10(10^4)

Since log10(10^4) = 4, we can simplify further:
pH = 4.76 + 0.058 + 4

Finally, adding up the numbers:
pH = 9.79

So, the pH of the buffer solution is approximately 9.79.