Question

If 50. 0 g of formic acid (HCHO2, ka = 1. 8 x 10^-4) and 30. 0 g of sodium formate (NaCHO2) are dissolved to make 500 ml. of solution, the ph of this solution is? a. 4. 76 b. 3. 76 c. 3. 35 d. 4. 12

Answer 1

If 50. 0 g of formic acid (HCHO2, ka = 1. 8 x 10^-4) and 30. 0 g of sodium formate (NaCHO2) are dissolved to make 500 ml. of solution, then the pH of this solution is 3.358.

Option C is correct option.

Given,

Given mass of formic acid = 50 g

Given mass of sodium formate = 30 g

Volume of formic acid = 500 ml

Volume of sodium formate = 500 ml

Molar mass of formic acid = 46 g

Molar mass of sodium formate = 68 g

Calculation of concentration of formic acid and sodium formate

Molar concentration is defined as the ratio of number of moles and the volume of solution.

Concentration of formic acid

Ca = 50/(46×500)

      = 0.00217 m

Concentration of sodium formate

Cb = 30/(68×500)

     = 0.00088 m

Using Henderson Hesselbalch equation,

pH = pKa + log(Cb/Ca)

pKa = -log(1.8 × 10^(-4))

      = 3.75

Substituting the value of pKa we get,

pH = 3.75 + log(0.00088/0.00217)

pH = 3.75-0.392

pH = 3.35

Thus the value of pH of solution containing formic acid and sodium formate is 3.35.

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