Question

Calculate the pH of a solution made by adding 35.0 g of sodium formate, NaHCOo, to 100 mL of 0.21 M formic acid, HCOOH. Hint what kind of solution is made? The K, for HCOOH is 1.8 x 10-4 M. As usual, report pH to 2 decimal places. Answer:

Answer 1

Molar mass of NaHCO2,

MM = 1*MM(Na) + 1*MM(H) + 1*MM(C) + 2*MM(O)

= 1*22.99 + 1*1.008 + 1*12.01 + 2*16.0

= 68.008 g/mol

mass(NaHCO2)= 35.0 g

use:

number of mol of NaHCO2,

n = mass of NaHCO2/molar mass of NaHCO2

=(35 g)/(68.01 g/mol)

= 0.5146 mol

volume , V = 1*10^2 mL

= 0.1 L

use:

Molarity,

M = number of mol / volume in L

= 0.5146/0.1

= 5.146 M

We have:

Ka = 1.8*10^-4

pKa = - log (Ka)

= - log(1.8*10^-4)

= 3.745

use:

pH = pKa + log {[conjugate base]/[acid]}

= 3.745+ log {5.146/0.21}

= 3.745+ 1.389

= 5.134

Answer: 5.13