for the titration of 10.0 ml of .15 M HNO2 aqueous solution (Ka=7.2x10^-4) using

Question

for the titration of 10.0 ml of .15 M HNO2 aqueous solution (Ka=7.2x10^-4) using 2.5 M NaOH aqueous solution, consider the following points of the titration: A: Initial pH of .15 M HNO2 aqueous solution B: .1 mL added base C: 1/2 equivalence point D: Equivalence point E Excess base (assume added 100.0 mL of base)

Set up the init mol, final mol, total volume, and ICE table for points B and D only:

Point B:

Titration equation:

init mol:

(assume reaction goes to completion)

final mol:

total volume:

flip reaction equation:

I:

C:

E:

Point D:

Titration equation:

init mol:

(assume reaction goes to completion)

final mol:

total volume:

flip reaction equation:

I:

C:

E:

Explanation / Answer

Point-B:

Titration equation: HNO2 + NaOH(aq) ----- > NaNO2(aq) + H2O

Initial moles of HNO2 = MxV(L) = 0.15 mol/L x 0.010 L = 0.0015 mol

Initial moles of NaOH during mixing = MxV(L) = 2.5 mol/L x 0.00010 L = 0.00025 mol

0.00025 mol of NaOH will react with 0.00025 mol of HNO2 to form 0.00025 mol of NaNO2(aq)

Hence

Final mol of HNO2 = 0.0015 mol - 0.00025 mol = 0.00125 mol

Final mol of NaOH = 0.00025 - 0.00025 = 0 mol

Total volume of solution = 10.0 mL+ 0.1 mL = 10.1 mL = 0.0101 L

--- HNO2 + NaOH(aq) ----- > NaNO2(aq) + H2O

I: 0.0015, 0.00025 mol, ----- 0 ------------ N.A

C: - 0.00025, - 0.00025 mol, + 0.00025, + 0.00025 mol

E: 0.00125 mol, 0 mol, ---------- +0.00025 mol, N.A

At equilibrium, [HNO2] = 0.00125 mol / 0.0101 L = 0.124 M

Point-D:

Titration equation: HNO2 + NaOH(aq) ----- > NaNO2(aq) + H2O

Initial moles of HNO2 = MxV(L) = 0.15 mol/L x 0.010 L = 0.0015 mol

Hence moles of NaOH required to reach the equivalence point = 0.0015 mol

=> MxV = 2.5 mol/L xV = 0.0015 mol

=> V = 0.0015 / 2.5 = 0.6 mL NaOH

Initial moles of NaOH during mixing = MxV(L) = 2.5 mol/L x 0.00060 L = 0.0015 mol

0.0015 mol of NaOH will react with 0.0015 mol of HNO2 to form 0.0015 mol of NaNO2(aq)

Hence

Final mol of HNO2 = 0.0015 mol - 0.0015 mol = 0.00 mol

Final mol of NaOH =0.0015 mol - 0.0015 mol = 0.00 mol

Total volume of solution = 10.0 mL+ 0.6 mL = 10.6 mL = 0.0106 L

--- HNO2 + NaOH(aq) ----- > NaNO2(aq) + H2O

I: 0.0015, 0.0015 mol, ----- 0 ------------ N.A

C: - 0.0015, - 0.0015 mol, + 0.0015, + 0.0015 mol

E: 0.00 , 0.00 mol, ---------- +0.0015 mol, N.A

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