The question is incomplete, here is the complete question:
Calculate the pH of a solution prepared by dissolving 0.370 mol of formic acid (HCO₂H) and 0.230 mol of sodium formate (NaCO₂H) in water sufficient to yield 1.00 L of solution. The Ka of formic acid is 1.77 × 10⁻⁴
a) 2.099
b) 10.463
c) 3.546
d) 2.307
e) 3.952
Answer: The pH of the solution is 3.546
Explanation:
We are given:
Moles of formic acid = 0.370 moles
Moles of sodium formate = 0.230 moles
Volume of solution = 1 L
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]
[tex]pH=pK_a+\log(\frac{[HCOONa]}{[HCOOH]})[/tex]
[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of formic acid = 3.75
[tex][HCOONa]=\frac{0.230}{1}[/tex]
[tex][HCOOH]=\frac{0.370}{1}[/tex]
pH = ?
Putting values in above equation, we get:
[tex]pH=3.75+\log(\frac{0.23/1}{0.37/1})\\\\pH=3.54[/tex]
Hence, the pH of the solution is 3.546
The pka of the formic acid is 3.546.
From the question, we are given
Moles of formic acid = 0.370 moles
Moles of sodium formate = 0.230 moles
Volume of solution = 1 L.
Molarity of a solution= moles of solute ÷volume.
The pH of the acidic buffer
pH=pka + log{salt/acid}
pH= pka +log {HCooNa/HCooH}
The pH of dissociation of formic acid is 3.75
Therefore, {HCooNa} =0.230/1
HCooH = 0.329/1
pH= 3.75+ log{0.23/1 ÷ 0.372/1}.
pH= 3.546.
pH is the degree of hydrogen ion or hydroxide ion in a solution.
Therefore, the pH is 3.546
The question is incomplete because the options are not given, here are the options from brainly website.
a.2.099
b) 10.463
c) 3.546
d) 2.307
e) 3.952
Learn more about pH from the link below.
https://brainly.com/question/13557815
