Question

To 1 L solution of 0.50 M HC2H3O2 and 0.50 M NaC2H3O2, 0.02 mol of solid NaOH was added. Calculate the pH of the resulting solution? Ka ( HC2H3O2) = 1.8 x 10-5. Use 3 sig.

Answer 1

mol of NaOH added = 0.02 mol

HC2H3O2 will react with OH- to form C2H3O2-

Before Reaction:
mol of C2H3O2- = 0.5 M *1.0 L
mol of C2H3O2- = 0.5 mol

mol of HC2H3O2 = 0.5 M *1.0 L
mol of HC2H3O2 = 0.5 mol

after reaction,
mol of C2H3O2- = mol present initially + mol added
mol of C2H3O2- = (0.5 + 0.02) mol
mol of C2H3O2- = 0.52 mol

mol of HC2H3O2 = mol present initially - mol added
mol of HC2H3O2 = (0.5 - 0.02) mol
mol of HC2H3O2 = 0.48 mol


Ka = 1.8*10^-5

pKa = - log (Ka)
= - log(1.8*10^-5)
= 4.745

since volume is both in numerator and denominator, we can use mol instead of concentration
use:
pH = pKa + log {[conjugate base]/[acid]}
= 4.745+ log {0.52/0.48}
= 4.779

Answer: 4.78

Answer 2

The balanced chemical equation for the reaction between HC2H3O2 and NaOH is:

HC2H3O2 + NaOH -> NaC2H3O2 + H2O

First, we need to determine the moles of HC2H3O2 and NaC2H3O2 in the solution before the addition of NaOH:

moles of HC2H3O2 = 0.50 M x 1 L = 0.50 mol moles of NaC2H3O2 = 0.50 M x 1 L = 0.50 mol

Since 0.02 mol of NaOH is added, it will react with HC2H3O2 and form NaC2H3O2. The remaining amount of HC2H3O2 and NaC2H3O2 will be:

moles of HC2H3O2 = 0.50 mol - 0.02 mol = 0.48 mol moles of NaC2H3O2 = 0.50 mol + 0.02 mol = 0.52 mol

Now, we can calculate the concentration of each species:

[HC2H3O2] = 0.48 L / 1 L = 0.48 M [NaC2H3O2] = 0.52 L / 1 L = 0.52 M

Next, we need to calculate the concentration of H+ ions produced from the dissociation of HC2H3O2:

Ka = [H+][C2H3O2-] / [HC2H3O2] 1.8 x 10^-5 = [H+] x [C2H3O2-] / 0.48 [H+] = sqrt(1.8 x 10^-5 x 0.48 / 1) = 1.59 x 10^-3 M

Now, we need to calculate the concentration of OH- ions produced from the reaction between NaOH and water:

NaOH -> Na+ + OH- [OH-] = moles of NaOH added / volume of solution [OH-] = 0.02 mol / 1 L = 0.02 M

Since NaOH is a strong base, it will react completely with the small amount of H+ ions produced from the dissociation of HC2H3O2. Therefore, the concentration of H+ ions will decrease by 0.02 M, while the concentration of Na+ and OH- ions will increase by 0.02 M each.

[H+] = 1.59 x 10^-3 M - 0.02 M = 1.39 x 10^-3 M [Na+] = [NaC2H3O2] = 0.52 M [OH-] = 0.02 M

Finally, we can calculate the pH of the resulting solution:

pH = -log[H+] pH = -log(1.39 x 10^-3) pH = 2.86

Therefore, the pH of the resulting solution is 2.86.