A 100.00 ml solution of 0.0530 M nitrous acid (Ka= 4.0 x 10^-4) is titrated with
ID: 552187 • Letter: A
Question
A 100.00 ml solution of 0.0530 M nitrous acid (Ka= 4.0 x 10^-4) is titrated with a 0.0515 M solution of sodium hydroxide as the titrant.
Part A) what is the pH of the acid solution after 15 mL of Titrant has been added ? (Kw= 1.00x10^-14) (answer pH= 2.63)
Part B) What is the pH of acid at equivalence point? (answer pH= 7.91)
Part C) What is the pH of the acid solution after 150.00 mL of the titrant has been added?(Kw= 1.00x10^-14) (answer pH= 11.99)
the answer for each is provided i just need the work behind it and why please!
Explanation / Answer
A 100.00 ml solution of 0.0530 M nitrous acid (Ka= 4.0 x 10^-4) is titrated with a 0.0515 M solution of sodium hydroxide as the titrant.
moles of nitrous acid = Molarirty* Volume in Liters= 0.053*100/1000 =0.0053
moles of KOH added = 0.0515*15/1000 = 0.000773
the reaction between HNO2( nitrous acid ) and KOH is HNO2+KOH------->KNO2+ H2O
molar ratio of HNO2: KOH (theoretical)= 1:1, actual molar ratio of HNO2 :KOH= 0.0053 :0.000773
so KOH is limiting reactants, all the KOH gets consumed. Moles of KNO2 formed =0.000773
moles of HNO2 remaining =moles of HNO2 present initially- moles of HNO2 consumed = 0.0053-0.000773=0.004528
volume after mixing = 100+15= 115ml= 115/1000L=0.115L
concentrations after mixing : HNO2= 0.004528/0.115 , KNO2= 0.000773/0.115
since pH= pKa+ log[A-]/[HA] where A- is KNO2 and [HA] is HNO2
Ka= 4*10-4, pKa= 3.4
pH= 3.4+ log [0.000773/0.004528)= 2.63
2. at Equivalance point of the reaction= moles of KOH= moels of HNO2= 0.0053
KOH required= 0.0053/0.0515 L=0.1029 L=102.9 ml
moles of KNO2 formed = 0.0053, volume of solution after mixing = 102.9+100 = 202.9 ml= 202.9/1000=0.203 L
concentration of KNO2= 0.0053/0.203 =0.026
KNO2 undergoes ionization as KNO2+ H2O--->NO2-+ OH- (1)
Kb= [NO2-] [OH-]/ [KNO2]
preparing the ICE Table
component Initial change Equilibrium
KNO2 0.026 -x 0.026-x
NO2- 0 x x
OH- 0 x x
where x= [OH-]
Kb= x2/(0.026-x) = 10-14/ Ka =10-14/ (4*10-4)= 2.5*10-11,
when solved for x using excel, x=8.07*10-7, pOH= -log (8.07*10-7)= 6.1
3. moles of KOH added = 0.0515*150/1000 =0.007725
now KOH is excess and all the HNO2 gets neutralized. moles of KOH additionally remaining after neutralization as per the reaction -1 given above, = 0.007725-0.0053=0.002425 moles
volume of reaction mixture= 150+100= 250ml =250/1000L=0.25L
concentration of remaining KOH= 0.002425/0.25=0.0097M
KOH being strong base ionizes completely into [OH-]
hence [OH-] =0.0097,pOH= 2.01, pH= 14-pOH= 14-2.01= 11.99
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