Place answers in space prowled and as much of your calculations as win fit addit
ID: 992778 • Letter: P
Question
Place answers in space prowled and as much of your calculations as win 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 "neutralize" 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-stoichiometric 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 hydrox.de required to reach the stoichiometric point has been added? Attach your graph to this test DON'T FORGET TO TITLE AND LABEL YOUR GRAPH Although not specified m parts a-f. you should have data points up to twice the volume required to reach the equivalence point to create a proper graph.Explanation / Answer
a) HNO2 + NaOH -------> NANO2 + H2O
HNO2 : NaOH = 1:1 moles
Required NaoH volume= V cm3
Reacted HNO2 moles = (0.125 ×20 )÷1000 mol
Reacted NaOH moles = (0.100 × V )÷1000 mol
Therefore consumed NaOH volume (V) = 25.00 cm3
b) Initial H+ concentration = 0.125 M
We know
pH = - log [H+]
So pH = -log [0.125]
pH = 0.9
c) at an half equilibrium,
Reacted HNO2 moles = (0.1×12.5 )÷1000 mol
= 1.25×10-3 mol
Initial aacid moles = (0.125 ×20)÷1000 = 2.5×10-3 mol
Remaining acid moles in the media = ( 2.5-1.25) ×10-3 mol
So H+ CONCENTRATION = (1.25 ×10-3 mol × 1000) ÷ 32.5 cm3
[H+]=0.0385 M
pH = -log (0.0385 )
pH =1.4