Answer
$pH = 9.66$
Work Step by Step
1000ml = 1L
25 ml = 0.025 L
25 ml = 0.025 L
1. Find the numbers of moles:
$C(HCl) * V(HCl) = 0.12* 0.025 = 3 \times 10^{-3}$ moles
$C(NH_3) * V(NH_3) = 0.43* 0.025 = 0.011$ moles
2. Write the acid-base reaction:
$HCl(aq) + NH_3(aq) -- \gt N{H_4}^+(aq) + Cl^-(aq)$
- Total volume: 0.025 + 0.025 = 0.05L
3. Since the acid is the limiting reactant, only $ 0.003$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[HCl] = 0.003 - 0.003 = 0$ mol.
$[NH_3] = 0.01075 - 0.003 = 7.8 \times 10^{-3} mol$
Concentration: $\frac{7.8 \times 10^{-3}}{ 0.05} = 0.16M$
$[N{H_4}^+] = 0 + 0.003 = 0.003$ moles.
Concentration: $\frac{ 0.003}{ 0.05} = 0.06M$
4. Since $N{H_4}^+$ is the conjugate acid of $NH_3$ , we can calculate its $K_a$ by using this equation:
$K_b * K_a = K_w = 10^{-14}$
$ 1.8\times 10^{- 5} * K_a = 10^{-14}$
$K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$
$K_a = 5.6\times 10^{- 10}$
5. Calculate the $pK_a$ for the acid
$pKa = -log(Ka)$
$pKa = -log( 5.6 \times 10^{- 10})$
$pKa = 9.25$
6. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 9.25 + log(\frac{0.16}{0.06})$
$pH = 9.25 + log(2.6)$
$pH = 9.25 + 0.41$
$pH = 9.66$
