Consider the titration of 100.0 mL of 0.200 M CH3NH2 by 0.100 M HCl. For each vo
ID: 876793 • Letter: C
Question
Consider the titration of 100.0 mL of 0.200 M CH3NH2 by 0.100 M HCl. For each volume of HCl added, decide which of the components is a major species after the HCl has reacted completely. Kb for CH3NH2 = 4.4 x 10- 4.Calculate the pH at the equivalence point for this titration?? The answer is NOT 10.65 or 10.64.
HINT: At the equivalence point, mol of base present initially exactly equals the mol of strong acid added. From this relationship, determine the volume of HCl necessary to reach the equivalence point. Next, determine the neutralization reaction then perform the stoichiometry calculation. From the major species present after the neutralization reaction, determine the initial concentration(s) of any species with acidic or basic properties then perform the equilibrium calculation to determine pH.
Consider the titration of 100.0 mL of 0.200 M CH3NH2 by 0.100 M HCl. For each volume of HCl added, decide which of the components is a major species after the HCl has reacted completely. Kb for CH3NH2 = 4.4 x 10- 4.
Calculate the pH at the equivalence point for this titration?? The answer is NOT 10.65 or 10.64.
HINT: At the equivalence point, mol of base present initially exactly equals the mol of strong acid added. From this relationship, determine the volume of HCl necessary to reach the equivalence point. Next, determine the neutralization reaction then perform the stoichiometry calculation. From the major species present after the neutralization reaction, determine the initial concentration(s) of any species with acidic or basic properties then perform the equilibrium calculation to determine pH.
Consider the titration of 100.0 mL of 0.200 M CH3NH2 by 0.100 M HCl. For each volume of HCl added, decide which of the components is a major species after the HCl has reacted completely. Kb for CH3NH2 = 4.4 x 10- 4.
Calculate the pH at the equivalence point for this titration?? The answer is NOT 10.65 or 10.64.
HINT: At the equivalence point, mol of base present initially exactly equals the mol of strong acid added. From this relationship, determine the volume of HCl necessary to reach the equivalence point. Next, determine the neutralization reaction then perform the stoichiometry calculation. From the major species present after the neutralization reaction, determine the initial concentration(s) of any species with acidic or basic properties then perform the equilibrium calculation to determine pH.
Explanation / Answer
Answer –
We are given, [CH3NH2] = 0.200 M , volume = 100.0 mL = 0.100 L
[HCl] = 0.100 M
Now we need to calculate pH at equivalence point –
We know at equivalence point moles of acid and moles of base are equal-
So first we need to calculate moles of given base
Moles of CH3NH2 = 0.200 M * 0.100 L
= 0.0200 moles
So, moles of HCl = 0.0200 moles
Now we need to calculate volume of HCl
Molarity = moles / volume (L)
Volume = moles/ molarity
= 0.0200 moles / 0.100 M
= 0.200 L
= 200 mL
Now reaction between acid and base
CH3NH2 + HCl -----> CH3NH3+ + Cl-
0.0200 0.0200 0.0200
So at equivalence point there is conjugate acid of base
Now we need to calculate new molarity of CH3NH3+
Total volume = 100 + 200 = 300 mL
= 0.300 L
[CH3NH3+] = 0.0200 mole / 0.300 L
= 0.0667 M
Now we need to put the ICE table
CH3NH3+ + H2O ------> CH3NH2 + H3O+
I 0.0667 0 0
C -x +x +x
E 0.0667-x +x +x
Ka = [CH3NH2] [H3O+] / [CH3NH3+]
Now we need to calculate Ka from Kb
We know, Ka * Kb = 1*10-14
Ka = 1*10-14 / 4.4*10-4
= 2.27*10-11
So, by plugging the values in it
2.27*10-11 = x* x / (0.0667-x)
The x in the 0.0667-x can be neglected, because Ka value is too small and it is also not obey 5% rule.
2.27*10-11 *0.0667 = x2
x2 = 1.515*10-12
x = 1.23*10-6 M
[H3O+] = x = 1.23*10-6 M
Now we know formula for calculate pH
pH = - log [H3O+]
= - log 1.23*10-6 M
= 5.91
pH at the equivalence point for this titration is 5.91