Questions 1: Consider the titration of a 24.0-mL sample of 0.175 M CH3NH2 with 0.155 M HBr. (The value of Kb for CH3NH2 is 4.4×10−4
A) Determine the initial pH
B) Determine the volume of added acid required to reach the equivalence point
C) Determine the pH at 4.0 mL of added acid
D) Determine the pH at one-half of the equivalence point.
E) Determine the pH at the equivalence point.
F) Determine the pH after adding 5.0 mL of acid beyond the equivalence point.
CH3NH2 = 24.0mL of 0.175M
HBr= 0.155M
Kb = 4.4x10^-4
a) Initial PH
Concentration of CH3NH2 = C= 0.175M
for weak bases
[OH-] = square root of Kb xC
[OH-]= square root of ( 4.4x10^-4 x 0.175)
[OH-]= 0.877 x10^-2
-log[OH-]= -log[0.877x10^-2]
POH = 2.06
PH + POH = 14
PH = 14 - POH
PH = 14 - 2.06
PH = 11.94
b) volume at equivalent
volume of HBr at equivalent point = 0.175 x 24.0 / 0.155 = 27.1 mL
volume of HBr = 27.1 mL
c) at the addition of 4 mL
CH3NH2 = 24.0mL of 0.175M
number of moles of CH3NH2 = 0.175M x 0.0240L = 0.0042 moles
HBr = 4.0mL of 0.155M
number of moles of HBr = 0.155M x 0.004L = 0.00062 moles
Kb = 4.4x10^-4
-log(Kb) = -log( 4.4x10^-4)
PKb = 3.36
CH3NH2 + HBr ----------------- CH3NH3+Br-
0.0042 0.00062 0
- 0.00062 - 0.00062 + 0.00062
0.00358 0 + 0.00062
POH = Pkb + log[salt]/[base]
POH = 3.36 + log( 0.00062/0.00358)
POH = 2.60
PH = 14 - POH
PH = 14 - 2.60
PH = 11.4
d) at one-half equivalent point
POH = PKb
POH = 3.36
PH = 14 - 3.36 = 10.64
PH = 10.64
e) at equivalent point
number of moles of CH3NH2 = 0.175M x 0.0240L = 0.0042 moles
HBr= 27.1mL of 0.155 moles
number of moles of HBr = 0.155M x 0.0271L = 0.0042 moles
at equivelent point, number of moles of CH3NH2 is equal to HBr
Total volume = 24.0+27.1 = 51.1ml = 0.0511L
Concentration = C = number of moles /Total volume = 0.0042/0.0511 = 0.0822 M
at equivalent point
PH = 7 - 1/2[PKb + logC]
PH = 7 - 1/2[3.36+ logC]
PH = 7 - 1/2[3.36 + log(0.0822)]
PH = 5.86
f) after adding 5 mL of acid beyond equivalent point
volume of Acid = 27.1 + 5.0 = 32.1 ml
number of moles of acid = 0.155M x 0.0321L = 0.00497 moles
number of moles of CH3NH2 = 0.175M x 0.0240L = 0.0042 moles
number of moles of acid is greater than the base
so the solution is acidic nature
remaining number of moles of acid = 0.00497 - 0.0042 = 0.00077 moles
Total volume = 24.0 + 32.1 = 56.1mL = 0.0561L
Concentration of H+ = number of moles/volume = 0.00077 / 0.0561 = 0.0137 M
[H+] = 0.0137M
-log[H+] = -log( 0.0137)
PH = 1.86.
Questions 1: Consider the titration of a 24.0-mL sample of 0.175 M CH3NH2 with 0.155 M...
Consider the titration of a 26.0 −mL sample of 0.180 M CH3NH2 with 0.155 M HBr. Determine each of the following. Part A: the initial pH Express your answer using two decimal places. Part B: the volume of added acid required to reach the equivalence point Part C: the pH at 6.0 mL of added acid Express your answer using two decimal places. Part D: the pH at one-half of the equivalence point Express your answer using two decimal places....
Consider the titration of a 27.0 −mL sample of 0.170 M CH3NH2 with 0.155 M HBr. Determine each of the following. Part A the initial pH Express your answer using two decimal places. pH = SubmitMy AnswersGive Up Part B the volume of added acid required to reach the equivalence point V = mL SubmitMy AnswersGive Up Part C the pH at 6.0 mL of added acid Express your answer using two decimal places. pH = SubmitMy AnswersGive Up Part...
Consider the titration of a 25.0 −mL sample of 0.180 M CH3NH2 with 0.150 M HBr. Determine each of the following: a) the initial pH b) the volume of added acid required to reach the equivalence point c) the pH at 4.0 mL of added acid d) the pH at one-half of the equivalence point e) the pH at the equivalence point f) the pH after adding 4.0 mLof acid beyond the equivalence point
Consider the titration of a 28.0 −mL sample of 0.170 M CH3NH2 with 0.145 M HBr. Determine each of the following. a) the initial ph b)the volume of added acid required to reach the equivalence point c)the pH at 4.0 mL of added acid d)the pH at one-half of the equivalence point e)the pH at the equivalence point f)the pH after adding 6.0 mL of acid beyond the equivalence point
Consider the titration of a 25.0 mL sample of 0.180 molL−1 CH3NH2 (Kb=4.4×10−4) with 0.155 molL−1 HBr. Determine the pH at 5.0 mL of added acid
consider the titration of a 34.0 mL sample of a 0.180 M HBr with 0.210 M KOH. determine the following: a. initial pH b. the volume if added base required to reach the equivalence point c. the pH at 10.6 mL of added base d. the pH at the equivalence point e. the pH after adding 5.0 mL of base beyond the equivalence point
consider the titration of a 25.7 mL sample of 0.115 M RbOH with 0.110 M HCl. Determine each of the following.a) the initial pHb) the volume of added acid required to reach the equivalence pointc) the pH at 4.4 mL of added acidd) the pH at the equivalence pointe) the pH after adding 5.2 mL of acid beyond the equivalence point
Consider the titration of a 25.1 −mL sample of 0.125 M RbOH with 0.100 M HCl. Determine each of the following.the initial pH, the volume of added acid required to reach the equivalence point,he pH at 4.9 mL of added acid,the pH at the equivalence pointthe pH after adding 4.2 mL of acid beyond the equivalence point
7. Consider the titration of a 33.0 mL sample of 0.170 M HBr with 0.200 M KOH. Determine each of the following: A. the initial pH B. the volume of added base required to reach the equivalence point (mL) C. the pH at 12.0 mL of added base (express your answer using three decimal places.) D. the pH at the equivalence point (express your answer as a whole number.) E. the pH after adding 5.0 mL of base beyond the...
Consider the titration of a 23.9 −mL sample of 0.125 MRbOH with 0.110 M HCl. Determine each of the following. a. the initial pH b. the volume of added acid required to reach the equivalence point c.the pH at 4.1 mL of added acid d.the pH at the equivalence point e.the pH after adding 5.1 mL of acid beyond the equivalence point