re 4/7/2017 11 PM 4.5ls G 4/7/2017 12:47 AM Gradebook Print Calculator Periodic

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re 4/7/2017 11 PM 4.5ls G 4/7/2017 12:47 AM Gradebook Print Calculator Periodic Table Question 3 of 14 Map Sapling Learning Calculate the pH of the solution after the addition of the following amounts of 0.0507 M HNO3 to a50.0 mL solution of 0.0750 M aziridine The pKa of azinidinium is 8.04. d) 69.2 mL of HNO3 a) 0.00 mL of HNO3 Number umber pH 6.93 pH 10.46 b) 5.85 mL of HNO3 e) Volume of HNO3 equal to the equivalence point Number Numb pH 7.14 pH 4.78 c) Volume of HNO3 equal to half 78.4 mL of HNO3 the equivalence point volume Number Number pH 8.04 4.77 pH ncorrect. Previous Try Again Next Ext Explanation

Explanation / Answer

This is a base, so better use pKb

pKb = 14-pKa = 14-8.04 = 5.96

b)

mmol of acid = MV =5.85*0.0507 = 0.296595 mmol of acid

mmol of base = MV = 50*0.075 = 3.75 mmol of base

mmol after reaction = 3.75-0.296595 = 3.4534

mmol of conjgute acid formed = 0.296595

this is a basic buffer so

pOH = pKb + log(BH+/B)

pOH = 5.96 + log(0.296595/3.4534 ) = 4.89391

pH = 14-pH = 14-4.89391

pH = 9.10609

then

f)

mmol of acid = MV =78.4*0.0507 = 3.97488 mmol of acid

mmol of base = MV = 50*0.075 = 3.75 mmol of base

there is clearly excess of acid

so

mmol of H+ = 3.97488 - 3.75 = 0.22488 mmol of H+

Vtotal = V1+V2 = 50+78.4 = 128.4

[H+] = mmol/Vtotal = 0.22488 / (128.4) = 0.001751401 M

pH = -log(0.001751401) = 2.75661

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