Calculate the pH of a 50.0 mL solution 0.0120 M ethylaminium chloride (C2H5NH3Cl ) and 0.0360 M ethylamine C2H5NH2. Kb = 4.3 x 10-4 Calculate the pH when 5.0 mL of 0.1 M HI is added. Ignore any changes in the volume
Use the Henderson-Hasselbalch equation.
To calculate the pH of the solution after adding 5.0 mL of 0.1 M HI (hydroiodic acid), we need to consider the acid-base reaction that occurs between HI and ethylamine.
The balanced equation for the reaction is:
HI(aq) + C2H5NH2(aq) -> C2H5NH3+(aq) + I-(aq)
Given that 5.0 mL of 0.1 M HI is added, the moles of HI added can be calculated as follows:
moles of HI = volume of HI (in L) * concentration of HI
= 5.0 mL * (1 L / 1000 mL) * 0.1 M
= 0.0005 mol
Since the reaction is 1:1, this means that 0.0005 mol of C2H5NH2 will react to produce 0.0005 mol of C2H5NH3+.
Now, let's calculate the new concentration of C2H5NH3+ after the reaction:
volume of C2H5NH3+ = initial volume of solution (50.0 mL) + volume of HI added (5.0 mL)
= 55.0 mL = 0.055 L
new concentration of C2H5NH3+ = moles of C2H5NH3+ / volume of C2H5NH3+
= 0.0005 mol / 0.055 L
= 0.00909 M
To find the pOH, we can use the Kb value:
Kb = [C2H5NH3+][OH-] / [C2H5NH2]
We know the value of Kb (4.3 x 10^(-4)), concentration of C2H5NH2 (0.0360 M), and the new concentration of C2H5NH3+ (0.00909 M). Rearranging the equation:
[OH-] = (Kb * [C2H5NH2]) / [C2H5NH3+]
= (4.3 x 10^(-4) * 0.0360) / 0.00909
= 0.00171 M
Now, let's calculate the pOH:
pOH = -log[OH-]
= -log(0.00171)
= 2.77
Finally, we can calculate the pH using the pOH:
pH = 14 - pOH
= 14 - 2.77
= 11.23
Therefore, the pH of the solution after adding 5.0 mL of 0.1 M HI is approximately 11.23.
To calculate the pH of the resulting solution after adding 5.0 mL of 0.1 M HI, we need to find the final concentrations of the ethylaminium chloride (C2H5NH3Cl), ethylamine (C2H5NH2), and hydronium ions (H3O+).
Here is how you can approach the problem:
1. Determine the initial concentration of the ethylaminium chloride (C2H5NH3Cl) and ethylamine (C2H5NH2):
- Concentration of C2H5NH3Cl = 0.0120 M
- Concentration of C2H5NH2 = 0.0360 M
2. Calculate the moles of ethylaminium chloride (C2H5NH3Cl) and ethylamine (C2H5NH2) based on the given volumes:
- Moles of C2H5NH3Cl = concentration (M) x volume (L)
Moles of C2H5NH3Cl = 0.0120 M x (50.0 mL / 1000 mL/L) = 0.0006 moles
- Moles of C2H5NH2 = concentration (M) x volume (L)
Moles of C2H5NH2 = 0.0360 M x (50.0 mL / 1000 mL/L) = 0.0018 moles
3. Write the balanced equation for the reaction between ethylaminium chloride (C2H5NH3Cl) and water (H2O):
C2H5NH3Cl + H2O ⇌ C2H5NH2 + H3O+
Kb = [C2H5NH2][H3O+] / [C2H5NH3Cl]
4. Calculate the concentration of hydronium ions (H3O+) at equilibrium using the Kb expression:
- Rearrange the equation: [H3O+] = Kb * [C2H5NH3Cl] / [C2H5NH2]
- Plug in the values: [H3O+] = (4.3 x 10^-4) * (0.0006) / (0.0018)
- Calculate: [H3O+] = 1.433 x 10^-4 M
5. Determine the concentration of hydronium ions (H3O+) after adding 5.0 mL of 0.1 M HI:
- Moles of HI = concentration (M) x volume (L)
Moles of HI = 0.1 M x (5.0 mL / 1000 mL/L) = 0.0005 moles
- The reaction between HI and water (H2O) is:
HI + H2O → H3O+ + I-
- Since the volume of the solution doesn't change, the concentration of H3O+ will increase due to the dissociation of HI. The final concentration of H3O+ is:
Final [H3O+] = [initial H3O+] + [HI]
Final [H3O+] = 1.433 x 10^-4 M + 0.0005 moles / 0.055 L
Final [H3O+] ≈ 1.52 x 10^-2 M
6. Calculate the pH using the concentration of hydronium ions (H3O+):
- pH = -log[H3O+]
- Plug in the value: pH = -log(1.52 x 10^-2)
- Calculate: pH ≈ 1.82
Therefore, the pH of the resulting solution after adding 5.0 mL of 0.1 M HI is approximately 1.82.