Suppose that you have 0.500 L of each of the following solutions, and an unlimit

Question

Suppose that you have 0.500 L of each of the following solutions, and an unlimited supply of water. K_b = 6.3x10^-5 for trimethylamine, (CH_3)_3 N and K_a = 1.9x10^-5 for hydrazoic acid, HN_3. 0.114 mol L^-1 (CH_3)_3 N(aq) 0.124 mol L^-1 HN_3(aq) 0.102 mol L^-1 HI(aq) 0.113 mol L^-1 KOH(aq) Provide simple instructions for preparing 1.00 L of a buffer solution having pH = 9.00 at 298 K. Focus on preparing the buffer that contains the highest possible concentrations of the active components. Your instructions should include the volumes of the solutions required.

Explanation / Answer

pH of basicbuffer = 14 - (pkb+log(salt(or)acid/base))

pkb of (CH3)3N = -logkb = -log(6.3*10^-5) = 4.2

let us consider preparation of 1 L of 0.01 M buffer with pH = 9

so that,

total no of mole of buffer = 1*0.1 = 0.01 mole

no of mole of (CH3)3N required = x

no of mole of HI required = 0.01-x

9 = 14-(4.2+log((0.01-x)/x)

   x = 0.00137 mole

no of mole of (CH3)3N required = x = 0.00137 mole

volume of (CH3)3N must take = n/M = 0.00137/0.114 = 0.012 L

                                                   = 12 ml

no of mole of HI required = 0.1-x = 0.01-0.00137 = 0.00863 mole

volume of HI must take = n/M = 0.00863/0.102 = 0.0846 L = 84.6 ml

take 12 ml of (CH3)3N , 84.6 ml HI and add enough of water to make solution up to 1 L .

so that we will get 1 L of 0.01 M buffer

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