A) Determine the pH of a 0.98 x 10^-2 mol L solution of hydrocyanic acid (HCN) Ka = 4.0 x 10^-10
B) What is the pH of a 0.243mol L solution of methylamine? (pKb for CH3NH2 = 3.30)
C) determine the pH of a buffer solution of Na2CO3 (pkb = 3.68, 0.125 M) and NaHco3 (0.35M)
A.
...........HCN ==> H^+ + CN^-
I.......0.0098.....0......0
C.........-x........x.....x
E......0.0098-x.....x.....x
Substitute the E line into the Ka expression and solve for x = (H^+) then convert to pH.
B.
.......CH3NH2 + HOH ==> CH3NH3^+ + OH^-
Set up an ICE chart as in A above, substitute into Kb and solve for x = OH^-, then convert to pH.
C.
Use the Henderson-Hasselbalch equation.
pH = pKa + log (base)/(acid)
You know pKb, convert to pKa.
You know M base (Na2CO3) and M acid(NaHCO3)
Post your work if you get stuck on any of these.
To determine the pH of a solution, we need to use the relevant acid dissociation constant (Ka or pKa) or base dissociation constant (Kb or pKb). We also need to know the concentration of the acid or base in the solution.
A) pH of a hydrocyanic acid (HCN) solution:
The given value is the Ka for hydrocyanic acid, Ka = 4.0 x 10^-10. We are also given the concentration of the solution, which is 0.98 x 10^-2 mol/L.
Step 1: Write the balanced chemical equation representing the dissociation of hydrocyanic acid in water:
HCN + H2O ⇌ H3O+ + CN-
Step 2: Set up the Ka expression for hydrocyanic acid:
Ka = [H3O+][CN-] / [HCN]
Step 3: Since the concentration of H3O+ is the same as the concentration of CN- (since HCN is a weak acid), we can represent it as x:
Ka = x^2 / (0.98 x 10^-2)
Step 4: Solve the equation for x to get the concentration of H3O+ (which corresponds to pH):
x = √(Ka * 0.98 x 10^-2)
Step 5: Calculate the pH using the concentration of H3O+
pH = -log[H3O+]
B) pH of a methylamine (CH3NH2) solution:
In this case, we are given the pKb for methylamine, pKb = 3.30, and the concentration of the solution, which is 0.243 mol/L.
Step 1: Convert the pKb value to Kb using the relation:
Kb = 10^(-pKb)
Step 2: Write the balanced chemical equation for the reaction of methylamine with water:
CH3NH2 + H2O ⇌ CH3NH3+ + OH-
Step 3: Set up the Kb expression for methylamine:
Kb = [CH3NH3+][OH-] / [CH3NH2]
Step 4: Since the concentration of OH- is the same as the concentration of CH3NH3+ (since methylamine is a weak base), we can represent it as x:
Kb = x^2 / (0.243)
Step 5: Solve the equation for x to get the concentration of OH-:
x = √(Kb * 0.243)
Step 6: Calculate the pOH using the concentration of OH-:
pOH = -log[OH-]
Step 7: Calculate the pH using the pOH:
pH = 14 - pOH
C) pH of a buffer solution of Na2CO3 and NaHCO3:
In this case, we are given the pKb for Na2CO3, which is 3.68, and the concentrations of Na2CO3 (0.125 M) and NaHCO3 (0.35 M).
Step 1: Convert the pKb value to Kb using the relation:
Kb = 10^(-pKb)
Step 2: We need to determine the concentration of OH- in the solution.
For the reaction: Na2CO3 + H2O ⇌ CO3^2- + 2OH-
Using the Kb expression for Na2CO3:
Kb = [CO3^2-][OH-]^2 / [Na2CO3]
Step 3: Since the concentration of CO3^2- and OH- is the same (due to the stoichiometry of the reaction), we can represent it as x:
Kb = x^3 / (0.125)
Step 4: Solve the equation for x to get the concentration of OH-:
x = ∛(Kb * 0.125)
Step 5: Calculate the pOH using the concentration of OH-:
pOH = -log[OH-]
Step 6: Calculate the pH using the pOH:
pH = 14 - pOH