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
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