## Chemistry and Chemical Reactivity (9th Edition)

$pH (Buffer\ solution)= 9.11$ $pH (Pure\ NH_3) = 11.17$ - The final pH is lower than the pH of the pure ammonia solution. (Because we added an acid).
$NH_4Cl$ 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: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 9.25 + log(\frac{0.12}{0.16})$ $pH = 9.25 + log(0.73)$ $pH = 9.25 + -0.136$ $pH = 9.11$ ------- 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: $pH + pOH = 14$ $pH + 2.83 = 14$ $pH = 11.17$