Answer
The pH of the buffer will be equal to $5.05$.
Work Step by Step
1. Calculate the molar mass of $NaOH$:
22.99* 1 + 16* 1 + 1.01* 1 = 40g/mol
2. Calculate the number of moles
- 80 mg = 0.080g
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 0.080}{ 40}$
$n(moles) = 2\times 10^{- 3}$
3. Find the concentration in mol/L:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
- 100 ml = 0.100 L
$ C(mol/L) = \frac{ 2\times 10^{- 3}}{ 0.100} $
$C(mol/L) = 0.02$
4. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 1.8 \times 10^{- 5})$
$pKa = 4.745$
5. Using the Henderson–Hasselbalch equation:
- Since we have added 0.02M of $NaOH$, which is a base, the concentration of the acid will decrease, and the concentration of the conjugate base will increase, both by 0.02M.
$pH = pKa + log(\frac{[Base] + 0.02}{[Acid] - 0.02})$
$pH = 4.745 + log(\frac{0.241 + 0.02}{0.150 - 0.02})$
$pH = 4.745 + log(\frac{0.261}{0.13})$
$pH = 4.745 + 0.3027$
$pH = 5.05$
