Answer
$pH (Buffer\ solution)= 9.11$
$pH (Pure\ NH_3) = 11.17$
- The final pH is lower than the pH of the pure ammonia solution. (Because we added an acid).
Work Step by Step
$NH_4Cl$ contains $N{H_4}^+$ (weak acid) and $Cl^-$ (neutral)
So, we just need to consider the ammonium ion when calculating the pH:
1. Since $N{H_4}^+$ is the conjugate acid of $N{H_3}$ , 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}$
$K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$
$K_a = 5.6\times 10^{- 10}$
2. Calculate the molar mass:
14.01* 1 + 1.01* 4 + 35.45* 1 ) = 53.5g/mol
3. Calculate the number of moles
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{2.2}{ 53.5}$
$n(moles) = 0.041$
4. Find the concentration in mol/L:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 0.041}{ 0.25} $
$C(mol/L) = 0.16$
5. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 5.6 \times 10^{- 10})$
$pKa = 9.25$
6. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{0.12}{0.16}$
- 0.73: It is.
7. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{0.12}{5.6 \times 10^{-10}} = 2.2\times 10^{8}$
- $ \frac{0.16}{5.6 \times 10^{-10}} = 3\times 10^{8}$
8. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 9.25 + log(\frac{0.12}{0.16})$
$pH = 9.25 + log(0.73)$
$pH = 9.25 + -0.136$
$pH = 9.11$
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The pure ammonia solution will have pH equal to:
9. We have these concentrations at equilibrium:
-$[OH^-] = [N{H_4}^+] = x$
-$[NH_3] = [NH_3]_{initial} - x = 0.12 - x$
For approximation, we consider: $[NH_3] = 0.12M$
10. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][N{H_4}^+]}{ [NH_3]}$
$Kb = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.12}$
$Kb = 1.8 \times 10^{- 5}= \frac{x^2}{ 0.12}$
$ 2.2 \times 10^{- 6} = x^2$
$x = 1.5 \times 10^{- 3}$
Percent ionization: $\frac{ 1.5 \times 10^{- 3}}{ 0.12} \times 100\% = 1.2\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [N{H_4}^+] = x = 1.5 \times 10^{- 3}M $
$[NH_3] \approx 0.12M$
11. Calculate the pOH:
$pOH = -log[OH^-]$
$pOH = -log( 1.5 \times 10^{- 3})$
$pOH = 2.83$
12. Find the pH:
$pH + pOH = 14$
$pH + 2.83 = 14$
$pH = 11.17$
