Calculate the pH of a 5.70×10-1 M aqueous solution of ethylamine hydrochloride (C2H5NH3Cl).
(For ethylamine, C2H5NH2, Kb = 5.60×10-4.)
C2H5NH3Cl is a salt which dissociates in aqueous solution as
C2H5NH3^+ + Cl^-, then the C2H4NH3^+ hydrolyzes with water as follows:
....................C2H5NH3^+ + H2O ==> C2H5NH2 + H3O^+
I....................0.571 M.............................0......................0
C.........................-x.................................x.......................x
E..................0.571-x................................x.......................x
Ka for C2H5NH3^+ = (Kw/Kb) = (C2H5NH2)(H3O)/(C2H5NH3^+)
Ka = (1E-14/5.60E-4) = (x)(x)/(0.571 - x)
Solve for x = (H3O^+) and convert to pH.
Post your work if you get stuck.
To calculate the pH of the solution, we need to determine the concentration of hydroxide ions (OH-) in the solution first.
Ethylamine hydrochloride (C2H5NH3Cl) is a salt formed by the reaction of ethylamine (C2H5NH2) with hydrochloric acid (HCl). In water, the salt will dissociate into its ions:
C2H5NH3+ (aq) + Cl- (aq)
Since ethylamine hydrochloride is a salt of a weak base (ethylamine) and a strong acid (HCl), the cation (C2H5NH3+) of the salt will not react with water and can be considered neutral.
Therefore, we can focus on the anion (Cl-) and its reaction with water. The chloride ion will not react with water and therefore does not affect the pH of the solution. It is the remaining hydroxide ions (OH-) from the water that will determine the pH.
Since we have a solution of a weak base, ethylamine (C2H5NH2), we can use the concept of Kb (base dissociation constant) to find the concentration of hydroxide ions (OH-) produced.
The Kb expression for ethylamine is:
Kb = [C2H5NH2+][OH-] / [C2H5NH2]
We can rearrange this equation to solve for [OH-]:
[OH-] = (Kb * [C2H5NH2]) / [C2H5NH2+]
Given that Kb of ethylamine is 5.60×10-4 and the concentration of ethylamine is 5.70×10-1 M, we can substitute these values into the expression:
[OH-] = (5.60×10-4 * 5.70×10-1) / (5.70×10-1) = 5.60×10-4 M
Now that we know the concentration of hydroxide ions, we can calculate the concentration of hydronium ions (H3O+) using the relationship:
[H3O+] * [OH-] = 1.0 × 10^-14 M^2 (from the ion product constant, Kw)
[H3O+] = (1.0 × 10^-14 M^2) / [OH-] = (1.0 × 10^-14 M^2) / (5.60×10-4 M) = 1.79×10-11 M
Finally, we can calculate the pH using the formula:
pH = -log10[H3O+]
pH = -log10(1.79×10-11) = 10.75
Therefore, the pH of the 5.70×10-1 M aqueous solution of ethylamine hydrochloride is approximately 10.75.
To calculate the pH of an aqueous solution of ethylamine hydrochloride (C2H5NH3Cl), you need to consider the reaction of ethylamine (C2H5NH2) with water to form its conjugate acid, ethylammonium ion (C2H5NH3+), and hydroxide ion (OH-). Then you can use the equilibrium constant expression to find the concentration of hydroxide ions and subsequently calculate the pH.
Step 1: Write the balanced equation for the reaction of ethylamine with water:
C2H5NH2 + H2O <-> C2H5NH3+ + OH-
Step 2: Write the equilibrium constant expression for this reaction:
Kw = [C2H5NH3+][OH-] / [C2H5NH2]
Step 3: Rearrange the equation to solve for [OH-]:
[OH-] = (Kw * [C2H5NH2]) / [C2H5NH3+]
Step 4: Substitute the given values into the equation:
Kw = 1.0 × 10^-14 (at 25°C)
[C2H5NH2] = 5.70 × 10^-1 M
[C2H5NH3+] = Since it is in the form of ethylamine hydrochloride, it will completely dissociate into ethylammonium ions and chloride ions. Therefore, [C2H5NH3+] = [C2H5NH3Cl] = 5.70 × 10^-1 M.
Step 5: Calculate [OH-]:
[OH-] = (1.0 × 10^-14 * 5.70 × 10^-1) / (5.70 × 10^-1)
[OH-] = 1.0 × 10^-14 M
Step 6: Calculate the pOH:
pOH = -log[OH-]
pOH = -log(1.0 × 10^-14)
pOH = 14
Step 7: Calculate the pH:
pH = 14 - pOH
pH = 14 - 14
pH = 0
Therefore, the pH of the 5.70×10^-1 M aqueous solution of ethylamine hydrochloride is 0.