Question

A 0.3654 g portion of pure formic acid is dissolved in 50.00 mL of water and is titrated with 0.1086 M NaOH. What pH is expected during this analysis when the titration is 0%, 25%, 50%, 75%, 100% and110% complete? Ka = 1.77 * 10^-4

Answer 1

Weight of formic acid = 0.3654 g, Volume of water = 50 mL, Molecular weight of formic acid = 46.03 g/mol

Number of moles of formic acid =0.3654/46.03 = 0.00793

Molarity(M) = Number of moles/Volume in Liters  

Molarity of formic acid = 0.158

Molarity of NaOH= 0.1086,

                    HCOOH + H2O …..H3O+ + HCOO-

Initial            0.158                         0              0       

C                     -x                                  x                 x

E                   0.158-x                          x                  x

K_a= (x)(x)/(0.158-x) = 1.77 x 10-4

X= 0.0053 M = [ H3O⁺]

PH=-log [ H3O⁺] = -log(0.0053) = 3-log(5.3)=3-0.7242 = 2.2758

25 % titration means 25 % of 0.00793moles of NaOH should be added. (Note total moles of Formic acid = 0.00793 moles)

Means 0.00198 moles of NaOH should be added.

Molarity of NaOH = 0.1086,

(M = n/V)

So the volume of given NaOH needs to be used = (n/M) =0.00198/0.1086 = 0.0182 = 18.2 mL

So 18.2 mL of 0.1086 M NaOH solution needs to be added to complete 25 % of titration

Hence the required moles of NaOH should be added is (n) = MV

(0.0182)[ (18.2)/1000] = 0.0003318

HCOOH       +                 OH-   …………..> HCOO- + H2O

0.00793                           0.000332             0     

(0.00793- 0.000332)        0                          0.000332                      

Now total volume of after adding NaOH to formic acid = 50 + 18.2 =68.2 mL

Hence concentration of final [HCOO-] = [0.000332]/[0.0682] = 0.00486 M

Final Concentration of [HCOOH] = [0.00793- 0.000332]/[0.0682]= 0.114 M

P^H = P^ka + log {[conjugate base =A-]/ [acid= HA]}

Ka= 1.77 x 10-4, P^Ka= 3.753

P^H = 3.753 + log[0.00486/0.114] = 3.753 + log [0.0426]=3.753-1.370 = 2.383

Follow the same procedure for 50%, 75%, 100% and110%.