Question

I'm having trouble with the calculations for my chemistry lab: Determination of an Acid Dissociation Constant, Ka Half-Neutralization

Molarity of acetic acid: 2 M Molarity of NaOH: 1 M

Volume of acetic acid: 25 mL

NaOH Solution

Final Buret Reading: 6.23 mL

Initial Buret Reading: 0.01 mL

Volume added: 6.22 mL

Total Volume of Solution: 250 mL

pH: 3.85

Caculating Ka

Initial number of moles HAn: ? OH^-: ?

Number of moles at equilibrium: HAn: ? An^-: ?

Equilibrium concentrations, mol L^-1 HAn: ? An^-: ? H₃O⁺: ?

Ka: ?

Your help and explanation would be very much appreciated, thanks!

Answer 1

Initially :

number of moles of HA are $\dfrac {2 \text { M} \times 25 \text { mL}}{1000 \text { mL/L}} = 0.05 \text { mol}$

number of moles of NaOH are $\dfrac {1 \text { M} \times 6.22 \text { mL}}{1000 \text { mL/L}} = 0.00622 \text { mol}$

Total volume is 250 mL or 0.25 L

At equilibrium :

pH is 3.85

$[H_3O^+] = 10^{-pH} = 10^{-3.85} = 0.0001413 \text { M}$

Number of moles of hydronium ions in 250 mL of solution is

$\dfrac {250 \text { mL}\times 0.0001413 \text { M}} {1000 \text { mL/L}} = 3.53 \times 10^{-5} \text { moles}$

Number of moles of HA at equilibrium are

$0.05-0.00622-3.53 \times 10^{-5} = 0.043745 \text { mol}$

Number of moles of    $A^-$ at equilibrium are

$0.00622 + 3.53 \times 10^{-5} = 0.006255 \text { mol}$

The acid dissociation constant

$K_a = \dfrac {[A^-][H_3O^+]}{[HA]} \\ K_a = \dfrac {\dfrac {0.006255}{0.250} \times 0.0001413}{\dfrac {0.043745}{0.250}} \\ K_a = 2.02 \times 10^{-5}$