A buffer that is 0.40 M base, B, and 0.25 M in its conjugate acid, BH+ has a pH

Question

A buffer that is 0.40 M base, B, and 0.25 M in its conjugate acid, BH+ has a pH of 8.88. What is the pH after 0.04 mol of HCl is added to 0.25 L of this buffer

Explanation / Answer

Concentrations: [B] = 0,4M = Cb [BH+] = 0,31M = Cs [HCl] = [H3O+] = 0,002mol / 0,25L = 0,008M You have to consider the reaction B + H2O BH(+) + OH(-) from which [OH-] = Kb * ( Cb / Cs ) You already have the concentration of hydroxyl ions: pH = 8,84 ----> pOH = 5,16 ----> [OH-] = 10^-5,16 = 6,918 * 10^-6M substituting above: Kb = [OH-] * ( Cs / Cb ) = 5,361 * 10^-6 Now, when you add HCl, all the hydronium ions are consumed to give BH(+) and pH remains virtually unchanged: B .........+ .........H3O(+) BH(+) ......+ ....H2O 0,4 ...................0,008 ............0,31 0,4 - 0,008 ...........0 ............0,31 + 0,008 Considering now the reaction B + H2O BH(+) + OH(-) you'll have [OH-] = Kb * ( 0,392 / 0,318 ) = 6,608 * 10^-6M ----> pOH = 5,18 ---> pH = 8,82

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