What is the pH at thestoichiometric point for the titration of 50 ml of 0.10 M m

Question

What is the pH at thestoichiometric point for the titration of 50 ml of 0.10 M methylamine CH3NH2 (aq) with 0.10 M perchloric acid HClO4(aq)? For methylamine (CH3NH2), Kb = 3.6 x 10^–4.

a) 5.93

b) 2.22

c) 7.00

d) 2.37

e) 5.78

Explanation / Answer

CH3NH2 + H+ ===>CH3NH3+ The reaction is 1:1. Thus the moles of starting amount will(base) =moles of acid moles of base=.1M=moles/..05                    =.005 moles of CH3NH2=Moles of HCLO4 .005 moles of HCLO4/volume=.10MHCLO4 Volume of HCLO4 required is .05L     CH3NH2 + H+ ===>CH3NH3+ Has to be be in moles since strong acid is present I    .005        .005              0 C -.005        -.005          +.005 E     0              0                .005 [CH3NH3+]=.005/.1(total volume of solution) =.05M CH3NH3+ + H20 ====> CH3NH2 + H+ .05               ----                0              0 -x                  --------         x              x .05-x           --------           x              x Ka=[x][x]/[.05-x] 2.77E-11=x^2/.05-x x=1.1768 -log(x)=5.929 thus choice A

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