Course X learning.com/ibiscms/mod/ibis/view.php?id-2502919 Gradeb 28/2016 11:55
ID: 943051 • Letter: C
Question
Course X learning.com/ibiscms/mod/ibis/view.php?id-2502919 Gradeb 28/2016 11:55 PM 3.8/102/25/2016 06:51 PM Print CalculatorPeriodic Table tion 8 of 13 Map Mapdo pling leaming Caloulate the pt for each of the following cases in the itration of 35.0 mL of 0.130 M NaOH(ag) with 0.130 M HI(aq). Number Note: Enter your answers with two decimal piaces (a) before addition of any H Number (b) after addition of 13.5 mL of H tb) aher adsin of 13 5mofD Number (c) after addition of 22.5 mL of H Number (d) after the addition of 35.0 mL of HI Number ater the adlion af 25 m of tD Number nafter the addition of 500 mL of HI () after the addition of 50.0 mL of HID tratio....pdf "Explanation / Answer
a)
pOH = -log(0.13) = 0.88605
pH = 14-pH = 14-0.88605= 13.11395
b)
V = 13.5 ml acid
mmol of base = MV = 35*0.13 = 4.55
mmol acid = MV = 0.13*13.5 = 1.755
mmol left = 4.55-1.755 = 2.795
VT = V1+V2 = 13.5+35 = 48.5
[OH-] = mmol/ml = 2.795/48.5 = 0.05762
pOH = -log(0.05762 = 1.239
pH = 14-pOH = 14-1.239
ph = 12.761
c)
V = 22.5 ml acid
mmol of base = MV = 35*0.13 = 4.55
mmol acid = MV = 0.13*22.5 = 2.925
mmol left = 4.55-2.925= 1.625
VT = V1+V2 = 22.5+35 = 57.5
[OH-] = mmol/ml = 1.625/57.5= 0.028260
pOH = -log(0.028260= 1.548827
pH = 14-pOH = 14-1.548827
ph = 12.45
d)
V = 35 ml acid
neutralization occurs,
pH = 7
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