50 mL of 0.250 M ethylenediamine solution was mixed with 12.5mL of 1.00 M HCl. WHat is the pH of the solution

The correct answer and explanation is:

Let’s analyze the problem step-by-step:

Given:

• Volume of ethylenediamine (EN) solution, VEN=50.0 mLV_{EN} = 50.0 \, mL
• Concentration of EN, CEN=0.250 MC_{EN} = 0.250 \, M
• Volume of HCl solution, VHCl=12.5 mLV_{HCl} = 12.5 \, mL
• Concentration of HCl, CHCl=1.00 MC_{HCl} = 1.00 \, M

Goal:

Find the pH of the final solution after mixing.

Step 1: Understand the chemistry

• Ethylenediamine (EN) is a weak base with two amine groups NH2\mathrm{NH_2}. It can accept protons from HCl.
• HCl is a strong acid, so it will fully dissociate into H+H^+ ions.
• When EN and HCl are mixed, HCl will protonate EN, forming the protonated ethylenediamine and Cl−Cl^-.
• This is essentially a buffer system made from the weak base (EN) and its conjugate acid (protonated EN).

Step 2: Calculate moles of EN and HCl

nEN=CEN×VEN=0.250 M×0.0500 L=0.0125 moln_{EN} = C_{EN} \times V_{EN} = 0.250 \, M \times 0.0500 \, L = 0.0125 \, mol nHCl=CHCl×VHCl=1.00 M×0.0125 L=0.0125 moln_{HCl} = C_{HCl} \times V_{HCl} = 1.00 \, M \times 0.0125 \, L = 0.0125 \, mol

Step 3: Determine the limiting reagent and the reaction

Since nEN=nHCl=0.0125 moln_{EN} = n_{HCl} = 0.0125 \, mol, they are in a 1:1 molar ratio.

The reaction: EN+HCl→ENH++Cl−\text{EN} + \text{HCl} \rightarrow \text{ENH}^+ + \text{Cl}^-

Because the moles of EN and HCl are equal, all the ethylenediamine is protonated, forming 0.01250.0125 mol of its conjugate acid, and no free EN remains.

Step 4: Calculate concentration of species in the total volume

Total volume: Vtotal=50.0 mL+12.5 mL=62.5 mL=0.0625 LV_{total} = 50.0 \, mL + 12.5 \, mL = 62.5 \, mL = 0.0625 \, L

Concentration of protonated EN (the conjugate acid): CENH+=0.0125 mol0.0625 L=0.20 MC_{ENH^+} = \frac{0.0125 \, mol}{0.0625 \, L} = 0.20 \, M

No free EN remains.

Step 5: Determine pH

Since all EN is converted to the conjugate acid ENH+ENH^+, the solution is acidic, dominated by the protonated species.

To find the pH, we need the acid dissociation constant KaK_a of the conjugate acid.

• Ethylenediamine has Kb≈7.9×10−4K_b \approx 7.9 \times 10^{-4}.
• Use Ka×Kb=Kw=1.0×10−14K_a \times K_b = K_w = 1.0 \times 10^{-14} at 25°C.

Calculate KaK_a: Ka=KwKb=1.0×10−147.9×10−4≈1.27×10−11K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{7.9 \times 10^{-4}} \approx 1.27 \times 10^{-11}

Step 6: Calculate pH using KaK_a

Using the KaK_a for ENH+ENH^+: Ka=[H+][EN][ENH+]K_a = \frac{[H^+][EN]}{[ENH^+]}

Initially, we only have ENH+ENH^+ at 0.20 M. ENEN and H+H^+ come from dissociation of ENH+ENH^+.

Let x=[H+]x = [H^+]: Ka=x20.20−x≈x20.20K_a = \frac{x^2}{0.20 – x} \approx \frac{x^2}{0.20}

(Assuming xx is very small compared to 0.20)

Solve for xx: x2=Ka×0.20=1.27×10−11×0.20=2.54×10−12x^2 = K_a \times 0.20 = 1.27 \times 10^{-11} \times 0.20 = 2.54 \times 10^{-12} x=2.54×10−12=1.59×10−6 Mx = \sqrt{2.54 \times 10^{-12}} = 1.59 \times 10^{-6} \, M

Step 7: Calculate pH

pH=−log⁡[H+]=−log⁡(1.59×10−6)≈5.80pH = -\log[H^+] = -\log(1.59 \times 10^{-6}) \approx 5.80

Final answer:

pH ≈ 5.80

Explanation summary:

You start with ethylenediamine, a weak base, and add a strong acid (HCl). Because the moles of acid and base are equal, the acid completely protonates the base, converting all EN to its conjugate acid ENH+ENH^+. The solution behaves like a weak acid solution, so the pH is calculated using the acid dissociation constant derived from KbK_b of EN.

The calculated pH (~5.80) is acidic, but not strongly so, because ethylenediamine is a weak base and its conjugate acid is a weak acid. This explains the mildly acidic pH after mixing.