Calculate the pH of a 1.19M sodium formate, HCOONa, solution.
Let's call formate ion for-. It will hydrolyze as follows:
.........for- + HOH ==> Hfor + OH^-
I........1.19.............0.....0
C.........-x..............x.....x
E.......1.19-x............x.....x
Kb for for- = (Kw/Ka for formic acid) = (x)(x)/(1.19-x)
Solve for x = OH^- and convert that to pH.
To calculate the pH of a sodium formate solution, we need to consider the dissociation of sodium formate in water.
The chemical equation for the dissociation of sodium formate is as follows:
HCOONa ⇌ HCOO- + Na+
Sodium formate (HCOONa) dissociates into the formate ion (HCOO-) and a sodium ion (Na+). The formate ion acts as a weak base and can hydrolyze with water to form hydroxide ions (OH-) that increase the pH of the solution.
Step 1: Write the equilibrium expression for the hydrolysis of formate ion (HCOO-):
HCOO- + H2O ⇌ HCOOH + OH-
Step 2: Set up the initial and equilibrium concentrations given the concentration of the sodium formate solution. In this case, the concentration of sodium formate (HCOONa) is 1.19 M. Therefore, we start with an initial concentration of 1.19 M for both the formate ion (HCOO-) and the hydroxide ion (OH-).
Step 3: Use the dissociation constant (Kb) for the hydrolysis reaction of the formate ion, which is 0.0000135.
Step 4: Use an ICE (Initial, Change, Equilibrium) table to determine the concentrations at equilibrium.
HCOO- + H2O ⇌ HCOOH + OH-
I 1.19 0 0 0
C -x +x +x +x
E 1.19-x x x x
Step 5: Substitute the equilibrium concentrations into the Kb expression:
Kb = [HCOOH] [OH-] / [HCOO-]
0.0000135 = x * x / (1.19 - x)
Step 6: Solve the equation for x. Since this is a weak base, we can assume that x is much smaller than 1.19, so we can approximate 1.19 - x as 1.19:
0.0000135 = x * x / 1.19
x^2 = 0.0000135 * 1.19
x = √(0.0000135 * 1.19)
x ≈ 0.003049
Step 7: Calculate the concentration of hydroxide ions (OH-) and use it to find the pOH:
[OH-] = x = 0.003049 M
pOH = -log[OH-] = -log(0.003049) ≈ 2.52
Step 8: Calculate the pH of the solution using the equation:
pH = 14 - pOH ≈ 14 - 2.52 ≈ 11.48
Therefore, the pH of a 1.19 M sodium formate solution is approximately 11.48.