calculate the pH at the equivalence point in the titration of 50 mL of .10M meth

Question

  calculate the pH at the equivalence point in the titration of 50 mL of .10M methylamine with a 2.0M HCL solution.

Explanation / Answer

CH3NH2 + HCl     => CH3NH3+   + Cl- 50 mL *(1L/1000mL) *0.10 mol CH3NH2/L = 0.005 molesCH3NH2 also 0.005 moles CH3NH3+ formed, and 0.005 moles HCl consumed (1:1stoichiometry) Volume HCl added: 0.005 mol HCl *(1L/2 mol HCl) = 0.0025 L Total volume = 0.0025 L + 50 mL *(1L/1000mL) = 0.0525 L [CH3NH3+] = 0.005 mol/0.0525 L = 0.0952 M               CH3NH3+         + H2O          CH3NH2        +    H3O+ initial       0.0952 change        -x                                                          x                     x equil       0.0952 -x                                                  x                      x Ka for CH3NH3+ is 10(-10.64) = 2.291 e -11 Ka = [CH3NH2][H3O+]/[CH3NH3+] = (x*x)/(0.0952 - x) assum 0.0952 - x ~ 0.0952   (0.0952 >>> x) Ka = 2.291 e-11 = x*x/(0.0952) x = (0.0952*2.291e-11)0.5 = 1.48 x 10-6 = [H3O+] pH = -log[H3O+] = -log(1.48e-6) = 5.83

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