Calculate the pH of the solution resulting from the addition of 75.0 mL of 0.15 M KOH to 35.0 mL of 0.20 M HCN (Ka (HCN) = 4.9 × 10–10).
To calculate the pH of the resulting solution, we first need to determine the concentration of each species in the solution after the reaction takes place. Here's how to do it:
Step 1: Calculate the number of moles of KOH and HCN:
Moles of KOH = volume (in liters) x concentration
= 0.075 L x 0.15 mol/L
= 0.01125 mol
Moles of HCN = volume (in liters) x concentration
= 0.035 L x 0.20 mol/L
= 0.007 mol
Step 2: Determine the limiting reactant:
The limiting reactant is the reactant that is completely consumed in the reaction. To find the limiting reactant, we compare the moles of reactants according to their stoichiometry. The balanced equation for the reaction is:
HCN + KOH → H2O + KCN
From the equation, we see that the stoichiometric ratio between HCN and KOH is 1:1. Since we have fewer moles of HCN (0.007 mol) compared to KOH (0.01125 mol), HCN is the limiting reactant.
Step 3: Determine the concentration of each species after the reaction:
After the reaction, all the HCN (0.007 mol) reacts with KOH in a 1:1 ratio, producing equal amounts of H2O and KCN.
Since the volume of the resulting solution is equal to the volume of the limiting reactant (0.035 L), the concentration of KCN and H2O will be:
C(KCN) = Moles(KCN)/Volume(solution) = 0.007 mol/0.035 L = 0.2 mol/L
C(H2O) = Moles(H2O)/Volume(solution) = 0.007 mol/0.035 L = 0.2 mol/L
The concentration of HCN will be zero since it is completely consumed in the reaction.
Step 4: Calculate the concentration of H3O+ ions (or H+) in the resulting solution:
Since the KCN formed is a salt, it is a strong electrolyte and will completely dissociate in water. For each KCN molecule that dissociates, a hydroxide ion (OH-) is consumed from the KOH added.
Thus, the concentration of OH- ions will be equal to the initial concentration of KOH, which is 0.15 M.
Since water also undergoes autoionization and produces equal amounts of H3O+ and OH- ions, the concentration of H3O+ ions will also be 0.15 M.
Step 5: Calculate the pOH and pH of the solution:
pOH = -log[OH-]
= -log[0.15]
≈ 0.82
pH + pOH = 14 (at 25°C)
pH = 14 - pOH
= 14 - 0.82
≈ 13.18
Therefore, the pH of the resulting solution is approximately 13.18.
To calculate the pH of the resulting solution, we need to determine the concentrations of H+ and OH- ions in the solution.
First, let's determine the moles of KOH and HCN in the solution:
Moles of KOH = volume of KOH solution (in L) × concentration of KOH solution
= 0.075 L × 0.15 mol/L
= 0.01125 mol
Moles of HCN = volume of HCN solution (in L) × concentration of HCN solution
= 0.035 L × 0.20 mol/L
= 0.007 mol
Since HCN is a weak acid, it will partially dissociate in water:
HCN + H2O ⇌ H3O+ + CN-
To determine the concentration of HCN that dissociates, we can use the equation for the Ka of HCN:
Ka = [H3O+][CN-] / [HCN]
Assuming x is the concentration of HCN that dissociates:
x × x / (0.007 - x) = 4.9 × 10–10
Since the dissociation is small compared to the initial concentration of HCN (0.007), we can approximate 0.007 - x ≈ 0.007:
x^2 / 0.007 = 4.9 × 10–10
Simplifying, we get:
x^2 = (4.9 × 10–10) × 0.007
x^2 = 3.43 × 10–12
x ≈ 5.85 × 10–6
Therefore, the concentration of H3O+ (H+) is approximately 5.85 × 10–6 M.
Next, let's calculate the concentration of OH- ions:
Since KOH is a strong base, it will completely dissociate in water to form KOH:
KOH → K+ + OH-
Since the moles of KOH are 0.01125 mol and the volume of the solution is 0.075 L:
Concentration of OH- = moles of KOH / volume of solution
= 0.01125 mol / 0.075 L
= 0.15 M
Now, let's calculate the concentration of H+ ions by using the equation for water self-ionization:
[H+][OH-] = 1.0 × 10^-14
Substituting the known values, we have:
(5.85 × 10^-6)(0.15) = 1.0 × 10^-14
Simplifying, we find:
[H+] = (1.0 × 10^-14) / (0.15 × 5.85 × 10^-6)
= 1.08 × 10^-4 M
Finally, we can calculate the pH of the solution using the equation:
pH = -log10[H+]
pH = -log10(1.08 × 10^-4)
= -(-3.967)
≈ 3.97
Therefore, the pH of the solution resulting from the addition of 75.0 mL of 0.15 M KOH to 35.0 mL of 0.20 M HCN is approximately 3.97.