Question

2. Weak Acid versus Strong Base Derive a titration curve for the titration of 50.0 mL of 0.100 M formic acid, HCHO2 (Ka 1.80 x 104) with 0.100 M N2OH. Calculate the pH for the following volumes of NaOH (0 mL, 10 mL, 25 mL, 40 mL, 50 mL, 55 mL, 60 mL). Volume of N2OH, in milliters pH (a) (b) (c) (d) (e) (f) (g) 0 10 25 40 50 55 60 pH at the equivalence point Specify your choice of indicator

Answer 1

Weak acid vs. Strong Base :

50.0 ml of 0.1 M HCOOH against 0.1 M NaOH

HCOOH (aq) + OH^- (aq) $\rightleftharpoons$ HCOO^-(aq) + H₂O (l)

Voulme of 0.1 M NaOH needed for equivalence point : 50.0 ml

Initial pH (  0 ml of NaOH) :  

HCOOH (aq) + H₂O (l) $\rightleftharpoons$ HCOO^-(aq) + H₃O⁺ (aq)

At eq. ~1 x x

Ka =  = (HCOO^- ) (H₃O⁺ ) /    (HCOOH ) =1.8*10^-4

(H₃O⁺ ) = (1.8*10^-4*0.100)^1/2 =4.24*10^-3

pH = -log (H₃O⁺ ) = 2.37

Before equivalence point :

Adding NaOH converts a portion of the formic acid to its conjugate base.

Any such solution containing comparable amounts of a weak acid , and its conjugate
weak base, is act as buffer. So, pH can be calculated following equation :

pH = pKa + log ( ( A^-) / (HA ) )   

- At 10 ml of NaOH :

Conc. of Unreacted (HCOOH) = (50 ml *0.100 M) - (10ml *0.100 M) / 60 ml = 0.067 M

Conc. of (HCOO-) = moles of NaOH aded / (total Vol.) = (10ml *0.100 M) / 60 ml = 0.017 M

pH = 3.74+ log ( ( 0.017 ) / (0.067) ) = 3.14

- At 25 ml of NaOH :

Conc. of Unreacted (HCOOH) = (50 ml *0.100 M) - (25ml *0.100 M) / 75 ml = 0.033 M

Conc. of (HCOO-) = moles of NaOH aded / (total Vol.) = (25ml *0.100 M) / 75 ml = 0.033 M

pH = 3.74+ log ( ( 0.033 ) / (0.033) ) = 3.74

- At 40 ml of NaOH :

Conc. of Unreacted (HCOOH) = (50 ml *0.100 M) - (40ml *0.100 M) / 90 ml = 0.011 M

Conc. of (HCOO-) = moles of NaOH aded / (total Vol.) = (40ml *0.100 M) / 90 ml = 0.044 M

pH = 3.74+ log ( ( 0.044 ) / (0.011) ) = 4.34

-At equivalence point (50 ml NaOH) :

At the equivalence point, moles of formic acid initially present and the moles of NaOH added are identical. Since their reaction effectively proceeds to completion, predominate ion in solution is formate, which is a weak base.

To calculate the pH we first determine the concentration of HCOO^-,

( HCOO-) = moles of weak acid added / total volume =  0.100 M * 50.0 ml / 100 ml = 0.05 M

A^- is a weak base, Kb for it = Kw / Ka = 10^-14 / (1.8*10^-4) = 5.56*10^-11

HCOO^-(aq) + H₂O (l) $\rightleftharpoons$  HCOOH (aq) + OH^- (aq)

Kb = (HCOOH ) (OH^- ) / (HCOO-)

[OH^–] = (Kb*   ( HCOO^-) ) ^1/2 = (5.56*10^-11*  0.05 ) ^1/2 = 1.67 *10^-6 M

pH = 14 -pOH = 14 - (-log[1.67*10^-6 ] = 8.22

-pH after equivalence point : After the equivalence point NaOH is present in excess, and the pH is determined by OH^- conc. :

(OH^- ) = Excess moles of NaOH added/ Total VOl.   

- At 55 ml of NaOH : (5 ml *0.100 M) / 105 ml = 0.0048 M

  pH = 14 -pOH = 14 - (-log[0.0048 ] = 11.67

- At 60 ml of NaOH : (10ml *0.100 M) / 110 ml = 0.0091 M

  pH = 14 -pOH = 14 - (-log[0.0091 ] = 11.95

Since, equivalence point pH is 8.22, thus , indicator choosen :

Best indicator is which pKa of that is close to equivalence point pH , thus we shall choose cresol red or thymol blue for better end point.