Calculate the pH of the solution that results from the mixture. 170 mL of 0.12 M C2H5NH2 with 285.0 mL of 0.22 M C2H5NH3Cl.

This is a buffer problem. Use the Henderson-Hasselbalch equation. C2H4NH2 is the base.

wha are the relative numbers of h3o+ and oh ions in an acidic, a basic, and a neutral solution?

To calculate the pH of the resulting solution, we need to consider the equilibrium of the weak base C2H5NH2 (ethylamine) and its conjugate acid C2H5NH3+ (ethylammonium ion).

First, let's write the balanced chemical equation for the reaction between the weak base and its conjugate acid:

C2H5NH2 + H2O ⇌ C2H5NH3+ + OH-

Now, let's determine the initial moles of each component in the solution:

For C2H5NH2:
0.12 M = 0.12 mol/L
moles of C2H5NH2 = 0.12 mol/L x 0.17 L = 0.02 mol

For C2H5NH3Cl:
0.22 M = 0.22 mol/L
moles of C2H5NH3Cl = 0.22 mol/L x 0.285 L = 0.0627 mol

Since we have a weak base and its conjugate acid, we know that the base will react with water to form OH- ions. In this case, the C2H5NH3+ (conjugate acid) will react with water to produce OH- ions.

Now, let's determine the moles of C2H5NH3+ produced during the reaction:

From the balanced equation, we can see that for every 1 mole of C2H5NH3+, 1 mole of OH- is produced.
Therefore, moles of OH- = moles of C2H5NH3+ = 0.0627 mol.

Next, we need to calculate the total volume of the resulting solution:
Total volume = volume of C2H5NH2 + volume of C2H5NH3Cl
Total volume = 0.17 L + 0.285 L = 0.455 L

Now, let's determine the concentration of OH- ions in the resulting solution:
OH- concentration = moles of OH- / total volume
OH- concentration = 0.0627 mol / 0.455 L = 0.138 M

Finally, we can calculate the pOH (negative logarithm of OH- concentration) of the solution:
pOH = -log[OH-]
pOH = -log(0.138) ≈ 0.86

To find the pH, we need to subtract the pOH from 14 (pH + pOH = 14):
pH = 14 - 0.86 ≈ 13.14

Therefore, the pH of the resulting solution is approximately 13.14.