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

Answer 1

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:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:  

$pH=pK_a+\log(\frac{[salt]}{[acid]})$

$pH=pK_a+\log(\frac{[HCOONa]}{[HCOOH]})$

$pK_a$ = negative logarithm of acid dissociation constant of formic acid = 3.75

$[HCOONa]=\frac{0.230}{1}$  

$[HCOOH]=\frac{0.370}{1}$

pH = ?  

Putting values in above equation, we get:  

$pH=3.75+\log(\frac{0.23/1}{0.37/1})\\\\pH=3.54$

Hence, the pH of the solution is 3.546