The concept used to solve this question is to calculate the at the equivalence point of the titration.
is the measure of hydrogen ion concentration. Lower the, higher is the hydrogen ion concentration and lower is the hydroxide ion concentration.
The formula to calculate the is as follows:
An equilibrium constant for the dissociation reaction is the equal to the product of the concentrations of the respective ions divided by the concentration of the undissociated molecule. It represents the extent of dissociation.
Consider a general reaction of acid, .
The equilibrium constant for the dissociation of is given as follows:
Consider a general reaction of base, .
The equilibrium constant for the dissociation of is given as follows:
The formula relating and is as follows:
An equivalence point in a titration is the point at which the moles of acid equals to the moles of base added.
The reaction between methyl amine and hydrochloric acid is as follows:
At equilibrium, moles of equal to the moles of .
At equivalence point, total volume is doubled and the concentration is halved.
So, initial concentration of
Construct an ICE table for the dissociation reaction of as follows:
Solve for x.
The at the equivalence point is calculated as follows:
Ans:The at the equivalence point is .
Determining pH at equivalence point in a titration of a weak base and strong acid.
Calculate the pH at the equivalence point for the titration of 0.250 M0.250 M methylamine (CH3NH2)(CH3NH2) with 0.250 M HCl.0.250 M HCl. The ?bKb of methylamine is 5.0×10−4.5.0×10−4. pH=pH=
Calculate the pH at the equivalence point for the titration of 0.110 M methylamine ( CH 3 NH 2 ) with 0.110 M HCl . The Kb of methylamine is 5.0×10^-4
CALCULATE THE PH AT THE EQUIVALENCE POINT FOR THE TITRATION OF 0.250M METHYLAMINE (CH3NH2) WITH 0.25 HCL. THE Kb OF METHYLAMINE IS 5.0X10-4. (HINT: THERE IS NO VALUE FOR VOLUME IN THE PROBLEM SO I AM NOT SURE HOW TO PROCEED! PLEASE HELP.)
Calculate the pH at the equivalence point in the titration of 60.0 mL of 0.140 M methylamine (Kb = 4.4 × 10−4) with 0.270 M HCl.
Calculate the pH at the equivalence point in the titration of 50 mL of 0.19 M methylamine (Kb = 4.3 ×10−4) with a 0.38 M HCl solution.
Calculate the pH at the equivalence point for the titration of 0.240 M methylamine ( CH 3 NH 2 ) with 0.240 M HCl . The K b of methylamine is 5.0 × 10 − 4 .
A. Match each type of titration to its pH at the equivalence point. Weak acid, strong base Strong acid, strong base Weak base, strong acid pH less than 7 pH equal to 7 pH greater than 7 B. A 56.0 mL volume of 0.25 M HBr is titrated with 0.50 M KOH. Calculate the pH after addition of 28.0 mL of KOH. C. Consider the titration of 50.0 mL of 0.20 M NH3 (Kb=1.8 x 10^-5) with 0.20 M HNO3....
Calculate the pH at the equivalence point for the titration of 0.220 M methylamine (CH_3 NH_2) with 0.220 M HCI. the K_b of methylamine is 5.0 times 10^-4. pH =
Enter your answer in the provided box. Calculate the pH at the equivalence point in the titration of 50 mL of 0.18 M methylamine ( Kb = 4.3 × 10−4 ) with a 0.36 M HCl solution.
What is the pH at the equivalence point of a weak base-strong acid titration if 20.00 mL of NaOCl requires 28.30 mL of 0.50 M HCl? Ka = 3.0 × 10-8 for HOCl. A. A) 0.30 B. B) 2.18 C. C) 6.76 D. D) 4.03 E. E)7.01 F. F) 8.92 G. G) none of these