What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous formic acid (HCOOH)requires 29.80 mL of 0.3567 mol L^-1 NaOH? K_a =1.8 × 10^-4 for formic acid. A)2.06 B)5.48 C)8.52 D)11.94

Question

Quizplus · Accepted Answer

At the equivalence point of a weak acid-strong base titration, the amount of strong base added is enough to completely neutralize the weak acid. This means that the amount of strong base added is equal to the amount of weak acid initially present.  
Using the given information:  
Number of moles of NaOH used = 0.3567 mol/L × 0.02980 L = 0.01062556 mol  
Number of moles of HCOOH initially present = 0.01062556 mol  
Now we can use the equilibrium expression for the ionization of formic acid:  
HCOOH + H2O ⇌ H3O+ + HCOO-  
Ka = [H3O+][HCOO-]/[HCOOH] = 1.8 × 10^-4  
At the equivalence point, [HCOOH] = [HCOO-], so we can substitute and simplify:  
Ka = [H3O+]^2 /[HCOOH]  
[H3O+]^2 = Ka[HCOOH] = 1.8 × 10^-4 × 0.01062556 mol / 0.025 L = 7.56 × 10^-5 mol/L  
[H3O+] = sqrt(7.56 × 10^-5) = 0.00869 mol/L  
pH = -log[H3O+] = 8.06  
Therefore, the closest choice is C) 8.52.

What Is the Approximate PH at the Equivalence Point of a Weak