Question

Calculate the hypothetical pH BEFORE acid is added for the titration of 50.00 mL of 0.0800 M Formic Acid (HCOOH, Ka= 1.80x104) with 0.1000 M NaOH at 25°C.

Answer 1

1. Ka of HCOOH = 1.80*10^-4

pH of weak acid HCOOH solution ,

HCOOH (aq) <---------> H⁺ +  HCOO^-

Ka = (H⁺) (HCOO^-) / (HCOOH) = 1.80*10^-4

(HCOOH) = 0.0800 M ;  (H⁺) = (HCOO^-)

(H⁺) = (1.80*10^-4 * 0.0800 M )^1/2 = 3.79*10^-3 M

pH = 2.42

pH of 0.1000 M NaOH solution : pH = 14 - pOH = 14 - (-log(0.1000 )) = 13

2. After addition of 15 ml of 0.1000 M NaOH :

HCOOH (aq) + NaOH (aq) <---------> HCOO^-Na⁺ (aq) + H₂O (l)

pH = pKa + log ( HCOO^-) /( HCOOH)

[HCOOH] = moles unreacted HCOOH /total volume

= (50 ml*0.0800 M) -(15ml*0.1000 M) /65 ml = 0.0385 M

[HCOO^-] = moles of NaOH added /total Vol. = (15ml*0.10 M)/ 65 ml = 0.0231 M

so, pH = 3.75 + log (0.0231) /( 0.0385) = 3.53

3. After addition of 40.50 ml of 0.1000 M NaOH :

HCOOH (aq) + NaOH (aq) <---------> HCOO^-Na⁺ (aq) + H₂O (l)

40.00 ml is equivalence point for this titration : 50 ml*0.0800 / 0.1000 = 40.00 ml

moles of HCOOH = moles of NaOH added

After the equivalence point NaOH is present in excess, and the pH is determined by it only,

Excess of [OH^-] = excess moles of NaOH added /total Vol.

= (0.50 ml *0.1000 M)/ 90.50 ml = 5.525*10^-4 M

pH = 14 - pOH = 14 - (-log( 5.525*10^-4 )) = 10.74