Answer
(a) $pH = 8.78$
(b) $pH = 1.66$
Work Step by Step
1000ml = 1L
10ml = 0.01 L
20ml = 0.02 L
1. Find the numbers of moles:
$C(HCl) * V(HCl) = 0.475* 0.01 = 4.75 \times 10^{-3}$ moles
$C(NH_3) * V(NH_3) = 0.318* 0.02 = 6.36 \times 10^{-3}$ moles
2. Write the acid-base reaction:
$HCl(aq) + NH_3(aq) -- \gt N{H_4}^+(aq) + Cl^-(aq)$
- Total volume: 0.01 + 0.02 = 0.03L
3. Since the acid is the limiting reactant, only $ 0.00475$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[HCl] = 0.00475 - 0.00475 = 0$ mol.
$[NH_3] = 0.00636 - 0.00475 = 1.61 \times 10^{-3} mol$
Concentration: $\frac{1.61 \times 10^{-3}}{ 0.03} = 0.0537M$
$[N{H_4}^+] = 0 + 0.00475 = 0.00475$ moles.
Concentration: $\frac{ 0.00475}{ 0.03} = 0.158M$
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}$
$K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$
$K_a = 5.56\times 10^{- 10}$
5. Calculate the $pK_a$ for the acid
$pKa = -log(Ka)$
$pKa = -log( 5.56 \times 10^{- 10})$
$pKa = 9.25$
6. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 9.25 + log(\frac{0.0537}{0.158})$
$pH = 9.25 + log(0.339)$
$pH = 9.25 + (-0.47)$
$pH = 8.78$
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(b)
1000ml = 1L
15ml = 0.015 L
20ml = 0.02 L
1. Find the numbers of moles:
$C(HCl) * V(HCl) = 0.475* 0.015 = 7.12 \times 10^{-3}$ moles
$C(NH_3) * V(NH_3) = 0.318* 0.02 = 6.36 \times 10^{-3}$ moles
2. Write the acid-base reaction:
$HCl(aq) + NH_3(aq) -- \gt N{H_4}^+(aq) + Cl^-(aq)$
- Total volume: 0.015 + 0.02 = 0.035L
Since the base is the limiting reactant, only $ 0.00636$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[NH_3] = 0.00636 - 0.00636 = 0M$.
$[HCl] = 0.007125 - 0.00636 = 7.65 \times 10^{-4}$ mol
Concentration: $\frac{7.65 \times 10^{-4}}{ 0.035} = 0.0219M$
$[N{H_4}^+] = 0 + 0.00636 = 0.00636$ moles.
Concentration: $\frac{ 0.00636}{ 0.035} = 0.182M$
- We have a strong and a weak acid. We can ignore the weak one
(because it has a very low $K_a$), and calculate the pH based only on the strong acid concentration:
$[H_3O^+] = [HCl] = 0.0219M$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 0.0219)$
$pH = 1.66$
