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.)
The reaction of CH3NH2 with HCl is
CH3NH2 + HCl ----- > CH3NH3+ + Cl-
1 mol-------- 1 mol------- 1 ml ------- 1 mol
At equivalence point all of the CH3NH2 are neutralized and there will be only CH3NH3+ and Cl-
Since equal moles of CH3NH2 and HCl react with each other, equal volume of 0.250 M HCl is required to titrate equal volume of CH3NH2.
Hence the volume of the solution doubles at equivalence point
Hence [CH3NH3+] = 0.250 *V / 2V = 0.125 M
At equivalence point, hydrolysis of CH3NH3+ will occur to form CH3NH2 and H3O+
--------------------CH3NH3+ ------- > CH3NH2 and H3O+; Ka = 10-14 / Kb = 10-14 / 5.0*10-4 = 2.0*10-11
Init.Conc(M): 0.125 --------------- 0 --------------- 0
Eqm.conc(M):0.125(1 - x)----- 0.125x --------- 0.125x
2.0*10-11 = 0.125x*0.125x / 0.125(1 - x)
=> x = 1.265*10-5
Hence [CH3NH2] = 0.125x = 0.125*1.265*10-5 = 1.58*10-6 M
[CH3NH3+] = 0.125(1 - x)
0.125 M
Now applying Hendersen equation
pOH = pKb + log[CH3NH3+] / [CH3NH2] = - log(5.0*10-4 ) + log (0.125 M / 1.58*10-6 M)
=> pOH = 8.20
=> pH = 14 - 8.20 = 5.8 (answer)
CALCULATE THE PH AT THE EQUIVALENCE POINT FOR THE TITRATION OF 0.250M METHYLAMINE (CH3NH2) WITH 0.25...
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