Question

For the titration of 10.0 mL of 0.15 M HNO₂ aqueous solution (K_a = 7.2 x 10^-4)using 2.5 M NaOH aqueous solution, consider the following points of titration:

B: 0.1 mL added base
D: equivalence point

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

Point B:
titration equation:

initial mol:

(assume reaction goes to completion)

final mol:

total volume:

flip reaction equation:

I:

C:

E:

Point D:
titration equation:

initial mol:

(assume reaction goes to completion)

final mol:

total volume:

flip reaction equation:

I:

C:

E:

Answer 1

Answer – We are given, [HNO₂] = 0.15 M , volume = 0.010 L ,

[NaOH] = 2.5 M, Ka = 7.2*10^-4

Part B) 0.1 mL of bases added

Initial moles of HNO₂ = 0.150 M * 0.010 L

                                       = 0.0015 moles

Initial moles of NaOH = 2.5 M * 0.0001 L

                                      = 0.00025 moles

Reaction –

HNO₂ + NaOH -------> H_­2O + NaNO₂

0.0015   0.00025                        0.00025

So limiting reactant is NaOH , so moles of NO₂^- for is 0.00025 moles

Moles after the reaction

Moles of HNO₂ = 0.0015 – 0.00025 = 0.00125 moles

Moles of NO₂^- = 0.00025 moles

Total volume = 10 +0.1 = 10.1 mL

So new molarity , [HNO₂] = 0.00125 moles / 0.0101 L = 0.124 M

[NO₂^-] = 0.00025 moles / 0.0101 L = 0.0247 M

ICE table –

HNO₂ + OH^- -------> NO₂^- + H₂O

I 0.124                    0.0247

C   -x                        +x        

E 0.124-x               0.0247-x  

(We can use the Henderson Hasselbalch equation for calculate pH.

pH = pKa + log [NO₂^-] / [HNO₂] )

Part D) at equivalence point

We know at the equivalence point moles of acid and base are equal, we calculated moles of acid in the part B.

Initial moles of HNO₂ = 0.150 M * 0.010 L

                                       = 0.0015 moles

So , moles of acid = moles of NaOH = 0.0015 moles

Now we need to calculate the volume of NaOH required

Volume of NaOH = moles of NaOH / molarity of NaOH

                               = 0.00150 moles / 2.5 M

                               = 0.0006 L

                              = 0.6 mL

So total volume = 10.0+0.6 = 10.6 mL

So at equivalence point all moles of acid reacted with base and form the conjugate base NO₂^-

So, moles of NO₂^- = 0.00150 moles

[NO₂^-] = 0.00150 moles / 0.0106 L

            = 0.142 M

ICE table –

NO₂^- + H₂O -------> HNO₂ + OH^-

I 0.142                      0            0

C   -x                       +x         +x

E 0.142-x                +x          +x

( For this one we need to use Kb and which is calculated form the Ka*Kb= 1.0*10^-14)