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

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

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:

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

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

pH = ?  

Putting values in above equation, we get:  

Hence, the pH of the solution is 3.546

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong