Calculate the pH of the buffer formed by mixing equal volumes [C_2H_5NH_2]= 1.68 M with [HClO_4]= 0.966 M . Kb=4.3×10−4.
I like to work in millimols. Let's just take 100 mL of each solution. To save space and typing let's call C2H5NH2 just RNH2.
millimols RNH2 = mL x M = 100 x 1.68 = 168
mmols HClO4 = 100 x 0.966 = 96.6
...RNH2 + HClO4 => RNH3^+ + ClO4^-
I..168.....0.........0.......0
add.......96.6................
C.-96.6.-96.6.......+96.6
E...71.4...0.........96.6
Substitute the E line into the Henderson-Hasselbalch equation and solve for pH. Note that you need to convert Kb to pKa. Two steps to do that.
pKb = -log Kb, then
pKb + pKa = 14. Solve for pKa.
To calculate the pH of the buffer, we need to consider the reaction that occurs when C2H5NH2 and HClO4 are mixed. The reaction can be represented as follows:
C2H5NH2 + HClO4 ⇌ C2H5NH3+ + ClO4-
In this reaction, C2H5NH2 acts as a base and HClO4 acts as an acid. Since we are mixing equal volumes, the concentrations of C2H5NH2 and HClO4 will be the same.
To understand how the pH of the buffer is calculated, we need to consider the equilibrium expression for C2H5NH2:
Kb = [C2H5NH3+][OH-] / [C2H5NH2]
Given that Kb = 4.3×10^-4, we can use this equilibrium expression to find [OH-]. However, we need to convert Kb to Ka since we need to calculate the pH, which is a measure of acidity. The relationship between Kb and Ka is:
Kb × Ka = Kw
where Kw is the water dissociation constant (1.0 × 10^-14 at 25°C).
Rearranging the equation, we get:
Ka = Kw / Kb = (1.0 × 10^-14) / (4.3 × 10^-4) = 2.33 × 10^-11
Now, we know that in the buffer solution, [H+] = [OH-] due to the equilibrium reaction. Therefore, we can write [OH-] as x and [C2H5NH2] as 1.68 M - x.
Given that [H+] = 10^-pH, the equilibrium expression for the acid dissociation constant (Ka) can be written as:
Ka = [H+][C2H5NH3+] / [C2H5NH2]
Substituting the known values, we get:
2.33 × 10^-11 = (1.0 × 10^-pH)(x) / (1.68 - x)
Solving this equation, we can find the value of x, which represents [OH-]. Since [OH-] = [H+], we can find the pOH and consequently the pH of the solution using the relationship:
pOH = -log10 [OH-]
pH = 14 - pOH
This method allows us to calculate the pH of the buffer solution formed by mixing C2H5NH2 and HClO4.