(b) 100.0 mL of 0.40 M C2H5NH2 (Kb = 5.6 multiplied by 10-4) titrated by 0.80 M

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Explanation / Answer

the solution remains same I got 50ml as when halfway point comes, out of 100 ml of 0.67ml HCl, exactly 50ml has been neutralized. so M1V1 = M2V2 where, M1 = molarity of HCl = 0.67 M V1 = volume of HCl neutralized at halfway = 50 ml M2 = molarity of NaOH = 0.34 M V2 = volume of NaOH consumed at halfway hence i got 98.53 ml. I am reposting the soln for you again: (b) C2H5NH2 + H2O ---------> (C2H5NH3^+) + OH- now, Kb is given using this we can calculate the initial conc of OH- ions it is equal to sqrt[(0.4*5.6e-4)] = 0.015 M so moles of OH- ions = 0.015*0.1 = 0.0015 moles at halfway, 0.00075 moles of OH- are still left so conc of OH- at halfway = moles/volume = 0.0075 M hence pOH = -log(0.0075) = 2.125 pH = 14-pOH = 11.875 at equivalence, all OH- is neutralized, and no H+ is also present so pH = 7 at equivalence (c) moles of H+ ions present initially = 0.67*0.1 = 0.067 M at halfway moles H+ present = 0.067/2 = 0.0335 conc of H+ ions = moles/vol = 0.335 to neutralize this 0.335 moles of H+ ions present in 50 ml we need (50*0.67)/0.34 = 98.53 ml NaOH total vol of soln at halfway = 198.53 ml now, 0.0335 moles of H+ ions are in 198.53 ml conc of H+ = 1.69e-4 M pH at halfway = -log(1.69e-4) = 3.773 at equivalence all H+ is consumed, so pH = 7

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