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, the ph of this solution is 3.35.

Therefore, option C is the correct option.

Given,

Given mass of sodium formate = 30 g

Given mass of formic acid = 50 g

Volume of sodium formate = 500 ml

Volume of formic acid = 500ml

Molar mass of sodium formate = 68 g

Molar mass of formic acid = 46 g

To calculate concentration of sodium formate and formic acid

The ratio of number of moles and the volume of solution is Molar concentration of substance.

Concentration of sodium formate

Cb = 30/(68×500)

= 0.00088m

Concentration of formic acid

Ca = 50/(46×500)

= 0.00217m

Now,

by using Henderson hesselbalch equation,

pH = pKa + log(Cb/Ca)

pKa = -log(1.8 × 10^(-4))

= 3.75

pH = 3.75 + log(0.00088/0.00217)

pH = 3.75 - 0.392

pH = 3.35

Thus, we calculated that the value of pH of solution of formic acid and sodium formate is 3.35.

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