A solution of 25mL of 0.5M c2h5nh2 is titrated with hydrochloric acid solution. During titration 5mL of .5M hcl is added dropwise. As volumes are additive. Calculate the pH of the resulting solution

millimols C2H5NH2 = mL x M = 25 x 0.5 = 12.5

mmols HCl = 5 mL x 0.5M = 2.5 HCl.
.......C2H5NH2 + HCl --> C2H5NH3^+ + Cl^-
I.......12.5.......0.........0......0
added............2.5................
C.......-2.5....-2.5........+2.5....2.5
E.......10.0......0..........2.5....2.5

This is a buffered solution. Use the Hendersen-Hasselbalch equation.

base = 10 mmols.
acid = 2.5 mmols

To calculate the pH of the resulting solution, we need to determine the concentrations of the base (C2H5NH2) and the acid (HCl) after the reaction is complete.

First, let's calculate the number of moles of C2H5NH2 present initially:
Moles of C2H5NH2 = Volume (in liters) × Concentration (in M)
= 25 mL × (0.5 M / 1000 mL)
= 0.0125 moles

Next, since the volume is additive, we can calculate the new volume of the solution after adding the HCl solution:
Total Volume = Initial Volume + Volume of HCl Solution
= 25 mL + 5 mL
= 30 mL

Now, let's calculate the number of moles of HCl added:
Moles of HCl = Volume (in liters) × Concentration (in M)
= 5 mL × (0.5 M / 1000 mL)
= 0.0025 moles

Since HCl is a strong acid, it completely dissociates in water, resulting in the same number of moles of H+ ions. Therefore, the moles of H+ ions added during the titration are also 0.0025 moles.

Now, let's calculate the total moles of C2H5NH2 and H+ ions after the reaction:
Total moles = Initial moles of C2H5NH2 - Moles of H+ ions added
= 0.0125 moles - 0.0025 moles
= 0.01 moles

Since the volume of the resulting solution is 30 mL, we can calculate the molarity (concentration) of the resulting solution:
Concentration (M) = Total moles / Total volume (in liters)
= 0.01 moles / (30 mL / 1000 mL)
= 0.333 M

Now, using the concentration of the resulting solution, we can calculate the pOH:
pOH = -log10(Concentration)
= -log10(0.333)
= 0.4774

Finally, since pH + pOH = 14 (at 25°C), we can calculate the pH:
pH = 14 - pOH
= 14 - 0.4774
= 13.5226

Therefore, the pH of the resulting solution is approximately 13.52.