Question

Identifying weak acid values

Data A student is trying to identify a weak acid by measuring the pH of partially neutralized weak acid solutions. In the lab the student is preparing three solutions by adding 5.00 ml, 10.00 ml, and 15.00 mL of a 0.200 M NaOH solution to 10.00 mL of 0.500 M weak acid solution. The solutions are labeled A, B and C below. The student diluted the solutions to 100.0 mL using the deionized water and measured the pH for each solution. The pH values are given in the table below. (A) eq [HA] eq 1/[A] K [H3O+] mL 0.200 M NaOH 5.00 Solution Α B C pH 3.21 3.66 4.01 10.00 15.00 You must include all data you collected or you analyzed, all your calculations, all tables,

Answer 1

Identifying values for Weak Acid:

Solution A :

pH = -log (H⁺) = 3.21 , we get (H⁺) = (H₃O⁺) =6.166*10^-4 M

• Added 5.00 ml. of 0.200 M NaOH ; final solution diluted to 100 ml = 0.01 M
• 10.00 ml of 0.500 acid ;final solution diluted to 100 ml = 0.05 M

weak acid have following equilibrium in water :

HA + H₂O$\rightleftharpoons$ H₃O⁺ + A^- ;    Ka = (H₃O⁺)(A^-)/ (HA)

Neutralization reaction : ​​​​​​​HA (aq) + OH^-(aq)   $\rightleftharpoons$ H₂O (l) + A^- (aq)

Since, for weak acid and strong base : pH = pKa + log( (A^-) / (​​​​​​​HA))

After addition NaOH and diluting solution (to 100 ml) the concentrations of HA and A^- at equilibrium are :   (​​​​​​​HA)eq =    0.05 M -0.01 M = 0.04 M

   (A^-)eq = moles of OH^- reacted = 0.01 M and 1/(A) = 100

From,  pH = pKa + log( (A^-) / (​​​​​​​HA))

pKa = 3.82 or Ka = 1.5*10^-4

Solution B :

pH = -log (H⁺) = 3.66 , we get (H⁺) = (H₃O⁺) =2.188*10^-4 M

• Added 10.00 ml. of 0.200 M NaOH ; final solution diluted to 100 ml = 0.02 M
• 10.00 ml of 0.500 acid ;final solution diluted to 100 ml = 0.05 M

weak acid have following equilibrium in water :

​​​​​​​HA + H₂O$\rightleftharpoons$ H₃O⁺ + A^- ;    Ka = (H₃O⁺)(A^-)/ (​​​​​​​HA)

Neutralization reaction : ​​​​​​​HA (aq) + OH^-(aq)   $\rightleftharpoons$ H₂O (l) + A^- (aq)

Since, for weak acid and strong base : pH = pKa + log( (A^-) / (​​​​​​​HA))

After addition NaOH and diluting solution (to 100 ml) the concentrations of HA and A^- at equilibrium are :   (​​​​​​​HA)eq =    0.05 M -0.02 M = 0.03 M

   (A^-)eq = moles of OH^- reacted = 0.02 M and 1/(A) = 50

From,  pH = pKa + log( (A^-) / (​​​​​​​HA))

pKa = 3.83 or Ka = 1.5*10^-4

Solution B :

pH = -log (H⁺) = 4.01 , we get (H⁺) = (H₃O⁺) =9.77*10^-5 M

• Added 15.00 ml. of 0.200 M NaOH ; final solution diluted to 100 ml = 0.03 M
• 10.00 ml of 0.500 acid ;final solution diluted to 100 ml = 0.05 M

weak acid have following equilibrium in water :

​​​​​​​HA + H₂O$\rightleftharpoons$ H₃O⁺ + A^- ;    Ka = (H₃O⁺)(A^-)/ (​​​​​​​HA)

Neutralization reaction : ​​​​​​​HA (aq) + OH^-(aq)   $\rightleftharpoons$ H₂O (l) + A^- (aq)

Since, for weak acid and strong base : pH = pKa + log( (A^-) / (​​​​​​​HA))

After addition NaOH and diluting solution (to 100 ml) the concentrations of HA and A^- at equilibrium are :   (​​​​​​​HA)eq =    0.05 M -0.03 M = 0.02 M

   (A^-)eq = moles of OH^- reacted = 0.03M and 1/(A) = 33.33

From,  pH = pKa + log( (A^-) / (​​​​​​​HA))

pKa = 3.83 or Ka = 1.5*10^-4