20.0 ml of 1.00 M formic acid are combined with 10.0 ml of 1.00 M sodium formate. Calculate the expected pH.
Ka for formic acid is 1.77 x 10^-4
Use the Henderson-Hasselbalch equation.
Convert Ka to pKa.
10.0 mL x 1.0 M sodium format is 10 millimols base.
20.0 mL x 1.00 M formic acid is 20 mmols of the acid. Plug and chug.
To calculate the expected pH in this scenario, we need to consider the acid-base reaction between formic acid (HCOOH) and sodium formate (NaHCOO).
The balanced chemical equation for this reaction is as follows:
HCOOH + HCOONa ⇌ HCOONa + H2O
Initially, we have 20.0 ml of 1.00 M formic acid and 10.0 ml of 1.00 M sodium formate. We need to calculate the moles of acid and base present in the mixture so that we can determine the excess or consumed moles of each component.
Step 1: Calculate the moles of formic acid (HCOOH):
Moles of HCOOH = volume (L) × molarity (mol/L)
Moles of HCOOH = 0.020 L × 1.00 mol/L = 0.020 mol
Step 2: Calculate the moles of sodium formate (HCOONa):
Moles of HCOONa = volume (L) × molarity (mol/L)
Moles of HCOONa = 0.010 L × 1.00 mol/L = 0.010 mol
Step 3: Determine the excess or consumed moles of each component:
Since the reaction is balanced, we can see that for every 1 mole of formic acid that reacts, 1 mole of sodium formate is formed. The reaction will consume all of the sodium formate and the remaining formic acid will be in excess.
Therefore, the consumed moles of sodium formate = 0.010 mol
The excess moles of formic acid = initial moles of formic acid - consumed moles of sodium formate
Excess moles of formic acid = 0.020 mol - 0.010 mol = 0.010 mol
Step 4: Calculate the final concentration of formic acid and its conjugate base (HCOO-) after the reaction:
HCOOH (formic acid) is a weak acid that partially ionizes in water. Let's assume x moles of HCOOH ionize.
[HCOOH]final = [HCOOH]initial - excess moles of formic acid
[HCOOH]final = 0.020 mol/L - 0.010 mol/L = 0.010 mol/L
[HCOO-]final = consumed moles of sodium formate
[HCOO-]final = 0.010 mol/L
Step 5: Calculate the ratio of [HCOO-]final to [HCOOH]final:
[HCOO-] / [HCOOH] = 0.010 mol/L / 0.010 mol/L = 1.00
Step 6: Calculate the pH using the Ka value for formic acid:
Ka = [HCOO-] × [H+] / [HCOOH]
Since [HCOO-] is equal to [HCOOH], we can substitute their values into the equation:
1.77 x 10^-4 = (1.0 × 10^-2) × (x^2) / (1.0 × 10^-2)
Simplifying the equation, we get:
x^2 = 1.77 x 10^-4
x = √ (1.77 x 10^-4)
x ≈ 4.21 x 10^-3
So, [H+] = 4.21 x 10^-3 mol/L
Finally, we can calculate the pH as follows:
pH = -log[H+]
pH = -log(4.21 x 10^-3)
pH ≈ 2.38
Therefore, the expected pH of the solution is approximately 2.38.