Answer
The $pH$ of this solution is about: $8.6$.
Work Step by Step
$NH_4NO_3:$
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:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 9.255 + log(\frac{0.15}{0.6433})$
$pH = 9.255 + -0.6323$
$pH = 8.623$
