A 0.1937-g sample of an unknownmonoprotic acid was dissolved in 34.1mL of water

Question

A 0.1937-g sample of an unknownmonoprotic acid was dissolved in 34.1mL of water and titrated with 0.0685M NaOH solution. The volume of base required to bring thesolution to the equivalence point was 15.6 mL. (a) Calculate the molar mass of the acid.
1 g/mol

(b) After 11.9 mL of base had beenadded during the titration, the pH was determined to be5.91. What is theKa of the unknown acid?
2 (a) Calculate the molar mass of the acid.
1 g/mol

(b) After 11.9 mL of base had beenadded during the titration, the pH was determined to be5.91. What is theKa of the unknown acid?
2

Explanation / Answer

HA + OH- = H2O + A- acid and OH- react in 1:1 ratio moles of base added: 15.6 mL *(1L/1000 mL) *(0.0685 mol OH-/L) =0.0010686 moles OH- moles HA: 0.0010686 moles originally in solution (OH- reacts withHA 1:1) Acid mass; 0.1937 grams molar mass = 0.1937 g/0.0010686 moles = 181 g/mol b) 11.9 mL *(1L/1000 mL)*(0.0685 mol OH-/L) = 0.00081515 moles OH- reacts with 0.0010686 moles of acid leaving 0.0010686 - 0.00081515 = 0.00025345 moles acid left in 34.1mL + 11.9 mL = 46 mL So concentration of acid immediately after reaction: [HA]_initial = 0.00025345 moles/0.046 L = 0.00551 M also you formed 0.00081515 moles of deprotonated acid, A- [A-]_initial = 0.00081515 mol/0.046 L = 0.0177 M ICE TABLE              HA                     A-   +          H+ initial      0.00551                 0.0177            0 change    -x                            x                 x equil     (0.00551 -x)         (0.0177 +x)        x Ka = [H+][A-]/[HA] = (x)*(0.0177 +x)/(0.00551-x) We know that pH = 5.91 = -log[H+] = -log(x) Thus x = 10^(-5.91) = 1.23 e-6 Sub this into Ka = (1.23e-6)*(0.0177 + 1.23e-6)/(0.00551 - 1.23e-6)= 3.96 e-6

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