University:
Seattle Pacific University
Course:
CHM 3372 | Organic Chemistry II
Academic year:

2022
Views:

386

Pages:

5
Author:

Nesbyenvkk

A 0.10 M solution of formic acid (HCHO2) has a pH of 2.38 at 25°C (a) Calculate Ka for formic acid at this temperature. pH = - log [H+] = 2.38 log [H+] =-2.38 [H+] = 10 -2.38 = 4.1686 E-3 HCOOH Initial H+ .1 M C - E (0.1- + 0 4.1686 E-3 M 4.1686 E-3) M 0 + 4.1686 E-3 M + 4.1686 E-3 M 4.1686 E-3 M 4.1686 E-3 M Taking the dissociation as negligible- Ka= ( )( (b) What percentage of the acid is ionized in this 0.10 M ) = 1.8 E-4 solution? % dissociation = (Concentration of H+/concentration of acid) x 100 = ( 4.1686 E-3/0.10 ) x 100 = 4. 16% HCOO- A 0.10 M solution of formic acid (HCHO2) has a pH of 2.50 at 25°C (a) Calculate Ka for formic acid at this temperature. pH = - log [H+] = 2.50 log [H+] =-2.50 [H+] = 10 -2.50 = 3.16227 E-3 HCOOH I C E H+ .1 M - + 0 3.16227 E-3 M (0.2- 3.16227 E-3) M 0 + 3.16227 E-3 M + 3.16227 E-3 M 3.16227 E-3 M 3.16227 E-3 M Taking the dissociation as negligible- Ka= ( )( ) = 1.0 x 10 -4 (b) What percentage of the acid is ionized in this 0.10 M solution? % dissociation = (Concentration of H+/concentration of acid) x 100 = ( 3.16227 E-3/0.10) x 100 = 3.16 % HCOO- A 0.10 M solution of formic acid (HCHO2) has a pH of 3.30 at 25°C a) Calculate Ka for formic acid at this temperature. pH = - log [H+] = 3.30 log [H+] =-3.30 [H+] = 10 -3.30 = 5.01187 E-4 HCOOH I C E H+ .1 M - + 0 5.01187 E-4 M 0 + 5.01187 E-4 M + 5.01187 E-4 M 5.01187 E-4 M 5.01187 E-4 M (0.3- 5.01187 E-4) M Taking the dissociation as negligible- Ka= ( )( HCOO- ) = 2.5 E-6 (b) What percentage of the acid is ionized in this 0.10 M solution? % dissociation = (Concentration of H+/concentration of acid) x 100 = ( 5.01187 E-4/0.10 ) x 100 = 0.50 % A 0.10 M solution of formic acid (HCHO2) has a pH of 3.2999999999999998at 25°C (a) Calculate Ka for formic acid at this temperature. pH = - log [H+] = 3.2999999999999998 log [H+] =-3.2999999999999998 [H+] 10 -3.2999999999999998 = 5.0118723362727209E-4 HCOOH .1 M - 5.0118723362727209E-4 M H+ + 0 HCOO0 + + 5.0118723362727209E- 5.0118723362727209E- 4M 4M 5.0118723362727209E- 5.0118723362727209E- 4M 4M (0.4- 5.0118723362727209E4) M Taking the dissociation as negligible- Ka= ( )( 2.511886431509578E-6 (b) What percentage of the acid is ionized in this 0.10 M solution? % dissociation = (Concentration of H+/concentration of acid) x 100 ) = = ( 5.0118723362727209E-4/0.10000000000000001) x 100 =

Formic Acid Key

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