Although the species of interest in this lab are phosphate ions, it is not possi

Question

Although the species of interest in this lab are phosphate ions, it is not possible to weigh out these ions separately. In this experiment, you will use their potassium salt as starting material. These salts are soluble in water and will completely dissociate to produce the desired phosphate ions. a. Determine the molecular weight of the salts below. b. Determine the pKa values of H_2PO^-_4 and HPO^2-_4. c. Predict qualitatively whether each salt will generate an acidic, basic or neutral solution when dissolved in water. a Determine the concentration of H_3O^+ ions in a solution which initially contains 0.050 mol/L of KH_2PO_4. State any assumption. b. What is the pH of this solution?

Explanation / Answer

Answer(2)

(a)

pKa = -log Ka

Since Ka is given to you, plug and chug. p always means - log of (e.g. pH = -log(H)

pKa = - log (Ka)

pKa = 7.21 is given so Ka = 6.2 x 10^-8

Kw=[OH-][H3O+]

6.2x10^-8=[0.050M][H3O+]

[H3O+]=1.24 x10^-6 M

(b)

Since pH = - log [H3O+]

pH = - log [1.24 x 10^-6]

pH = 5.90

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.