Please help!! and please show your work. A 15.00 mL sample of hydrofluoric acid,
ID: 1011766 • Letter: P
Question
Please help!! and please show your work.
A 15.00 mL sample of hydrofluoric acid, K_a = 6.8 Times 10^-4, was titrated with 0, 475 M KOH. The equivalence point was reached after addition of 19.35 mL of base. Determine the molar concentration of the original hydrofluoric acid solution, and find the pH of the original hydrofluoric acid solution. Phosphoric acid, a triprotic acid, is a common ingredient in soft drinks. The lionization equations for this acid are shown below: H_3PO_4 + H_2O H_2PO^-_4 + H_3O^+ K_a1 = 6.9 Times 10^-3 H_3PO^-_4 + H_2O HPO^2-_4 + H_3O^1+ K_a2 = 6.2 Times 10^8 HPO^2-_4 + H_2O PO^3-_4 + H_2O^- K_a3 = 4.8 Times 10^-13 The following equations represent the reactions that would take place during titration of phosphor acid with a strong base. Using K_and the K_values above, determine the equilibrium constants associated with each of following reactions: H_3PO_4 + OH^1- H_2PO^2-_4 + H_2O k^= ? H_2PO^1-_4 + OH^1- HPO^2-_4 + H_2O k^= ? HPO^2-_4 + OH^1- PO^3_4 + H_2O K^= ? Would you expect to see three equivalence points in the titration curve for phosphoric acid? ExplainExplanation / Answer
2. Titration of HF with KOH
moles of HF present = moles of KOH used to reach Eq. point = 0.475 M x 0.01935 L = 0.0092 mol
Molar concentration of HF solution = 0.0092 mol/0.015 L = 0.613 M
pH of HF solution
HF + H2O <==> F- + H3O+
let x amount of HF has dissociated
Ka = [F-][H3O+]/[HF] = 6.8 x 10^-4 = x^2/0.613
x = [H3O+] = 0.0204 M
pH = -log[H3O+] = 1.69
3. For H3PO4
H3PO4 + OH- <==> H2PO4- + H2O ...... K' = Ka1 = 6.9 x 10^-3
H2PO4- + OH- <==> HPO4^2- + H2O ....K'' = Ka2 = 6.2 x 10^-8
HPO4^2- + OH- <==> PO4^3- + H2O ....K''' = Ka3 = 4.8 x 10^-13
Yes we would expect three distinct equivalence points in the titration curve for H3PO4 with strong base KOH.