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!

Answer 1

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