## Chemistry: The Molecular Science (5th Edition)

The pH of the buffer will be equal to $5.05$.
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. $pH = pKa + log(\frac{[Base] + 0.02}{[Acid] - 0.02})$ $pH = 4.745 + log(\frac{0.241 + 0.02}{0.150 - 0.02})$ $pH = 4.745 + log(\frac{0.261}{0.13})$ $pH = 4.745 + 0.3027$ $pH = 5.05$