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

The $pH$ of this solution is about: $8.6$.
$NH_4NO_3:$ 1. Calculate the molar mass : 14.01* 1 + 1.01* 4 + 14.01* 1 + 16* 3 = 80.06g/mol 2. Calculate the number of moles $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 5.15}{ 80.06}$ $n(moles) = 0.06433$ 3. Find the concentration in mol/L: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 0.06433}{ 0.1}$ $C(mol/L) = 0.6433$ 4. Since $N{H_4}^+$ is the conjugate acid of $NH_3$ , we can calculate its ka by using this equation: $K_b * K_a = K_w = 10^{-14}$ $1.8\times 10^{- 5} * K_a = 10^{-14}$ $K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$ $K_a = 5.556\times 10^{- 10}$ 5. Calculate the pKa Value $pKa = -log(Ka)$ $pKa = -log( 5.556 \times 10^{- 10})$ $pKa = 9.255$ 6. Check if the ratio is between 0.1 and 10: - $\frac{[Base]}{[Acid]} = \frac{0.15}{0.6433}$ - 0.2332: It is. 7. Check if the compounds exceed the $K_a$ by 100 times or more: - $\frac{0.15}{5.556 \times 10^{-10}} = 2.7\times 10^{8}$ - $\frac{0.6433}{5.556 \times 10^{-10}} = 1.157\times 10^{9}$ 8. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 9.255 + log(\frac{0.15}{0.6433})$ $pH = 9.255 + -0.6323$ $pH = 8.623$