Place answers in space provided and as much of your calculations as will fit, ad
ID: 993188 • Letter: P
Question
Place answers in space provided and as much of your calculations as will fit, additional pages are available on line For the titration of 20.0 mL of 0.125 M nitrous acid HNO_2 with 0.100 M sodium hydroxide, NaOH. answer the following questions and construct a titration curve (Please use pK_a = 3.34 for nitrous acid) What volume (in ml) of 0 100M sodium hydroxide is required to Prime neutralize Prime 20.0 mL of 0 125M nitrous acid? What is the pH of the solution before any sodium hydroxide is added? What is the pH at the half-stoichiometnc or half-equivalence point (that is what is the pH when you have added 1/2 of the total volume of base that would be required to react with all of the acid originally present)? What is the pH when 95% of the sodium hydroxide required to reach the stoichiometric point has been added? What is the pH at the stoichiometric point? What is the pH when a 5% excess of the sodium hydroxide required to reach the stoichiometric point has been added? Attach your graph to this test DON prime T FORGET TO TITLE AND LABEL YOUR GRAPH! Although not specified in parts a-f,you should have data points up to twice the volume required to reach the equivalence point toExplanation / Answer
a) Use the formula: MaVa = MbVb
Vb = 0.125 * 20 / 0.1 = 25 mL
b) pH before NaOH is added we have in solution the following:
r: HNO2 <----------> H+ + NO2- Ka = 10-3.34 = 4.57x10-4
i: 0.125 0 0
e: 0.125-x x x
4.57x10-4 = x2 / 0.125-x --> Ka is smaller, so we can approximate 0.125-x to 0.125 only:
4.57x10-4 * 0.125 = x2
x = [H+] = 7.56x10-3 M
pH = -log(7.56x10-3) = 2.12
c) at the half equivalence point (12.5 mL) the half moles are used, but these half moles are produce, so the ratio of the product and reactant will be 1. Therefore pH = pKa = 3.44
d) at 95% of NaOH ---> 25 * 0.95 = 23.75 mL
moles NaOH = 0.1 * 0.02375 = 2.375x10-3 moles
moles acid = 0.125 * 0.020 = 2.5x10-3 moles
remaining moles of acid = 2.5x10-3 - 2.375x10-3 = 1.25x10-4 moles
pH = 3.44 + log(2.375x10-3 / 1.25x10-4)
pH = 4.72
e) At the equivalence point, the moles of acid are consumed and the moles of base begins to be in excess so the species present in solution are:
moles present = 2.5x10-3 moles
Concentration = 2.5x10-3 / 0.045 = 0.055 M
r: NO2- + H2O <-------> OH- + HNO2 Kb = 1x10-14 / 4.57x10-4 = 2.19x10-11
i: 0.055 0 0
e: 0.055-x x x
2.19x10-11 = x2/0.055-x
2.19x10-11 * 0.055 = x2
x = [OH-] = 1.097x10-6 M
pOH = -log(1.097x10-6) = 5.96
pH = 14-5.96 = 8.04
Now with this guide, try to do part f) yourself and construct the graph.
Hope this helps