Question

6. (10 points) How many grams (g) of formic acid (HCHO2) would you need to make a 555mL solution with a pH of 2.25? Weah accu ossi

!!!!!!!!!!!!! KaHCOOH formic acid: 1.7 x 10^-4 !!!!!!!!!!!!!!!!

Answer 1

Let the concentration of HCOOH be c

use:

pH = -log [H+]

2.25 = -log [H+]

[H+] = 5.623*10^-3 M

HCOOH dissociates as:

HCOOH -----> H+ + HCOO-

c 0 0

c-x x x

Ka = [H+][HCOO-]/[HCOOH]

Ka = x*x/(c-x)

1.7*10^-4 = 5.623*10^-3*5.623*10^-3/(c-5.623*10^-3)

c-5.623*10^-3 = 0.186

c=0.1916 M

This is concentration of HCOOH

volume , V = 5.55*10^2 mL

= 0.555 L

use:

number of mol,

n = Molarity * Volume

= 0.1916*0.555

= 0.1063 mol

Molar mass of CH2O2,

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

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

= 46.026 g/mol

use:

mass of CH2O2,

m = number of mol * molar mass

= 0.1063 mol * 46.03 g/mol

= 4.894 g

Answer: 4.89 g