Salts form acidic or basic solutions through the same ionization equations of their conjugate acids and bases. However,

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Salts Form Acidic Or Basic Solutions Through The Same Ionization Equations Of Their Conjugate Acids And Bases However 1 (93.9 KiB) Viewed 12 times

Salts form acidic or basic solutions through the same ionization equations of their conjugate acids and bases. However, since the conjugate is the starting material, the corresponding acid or base will form. For example, a solution of NH4Cl will ionize in solution to form NH4+ and Cl. The conjugate of NH3 will form some of the base through the equation NH4+ (aq) + H₂O(1) ⇒ NH3(aq) + H3O+ (aq) and thus NH4+ acts as a weak acid. Since Cl- is the conjugate of a strong acid, HCl, that completely ionizes in solution, no significant amount of OH will form, and the salt NH4Cl yields an acidic solution. When working with a polyprotic acid, at the first equivalence point, there will be just enough base to convert a polyprotic acid to its first salt. There will be an equivalence point as each proton is ionized in turn until only a salt with no ionizable protons remain. For example, the acid H₂A titrated with the base MOH would have a first equivalence at MHA, and a second at M₂ A. The pH of the salt where an ionizable proton remains, MHA, can be approximated using the equation pH at equivalence point with ionizable proton = PK₂1 +PK₂2 2 At 25 °C phosphoric acid, H3PO4, has the following equilibrium constants: H3PO4 (aq) + H₂O(1) = H3O+ (aq) + H₂PO2 (aq) H₂PO4 (aq) + H₂O(1) = H₂O+ (aq) + HPO4²- (aq) HPO4²- (aq) + H₂O(l) = H₂O+(aq) + PO4³- (aq) 3- Part A What is the pH of a solution of 0.25 MK3PO4, potassium phosphate? Express your answer numerically to the hundredths place. ► View Available Hint(s) 15. ΑΣΦ Bw ? pH = Submit Part B Kal Ka2 Ka3 = = = 7.5 × 10-3 6.2 × 10-8 4.2 x 10-13
Salts form acidic or basic solutions through the same ionization equations of their conjugate acids and bases. However, since the conjugate is the starting material, the corresponding acid or base will form. For example, a solution of NH4Cl will ionize in solution to form NH4+ and Cl. The conjugate of NH3 will form some of the base through the equation NH4+ (aq) + H₂O(1) ⇒ NH3(aq) + H3O+ (aq) and thus NH4+ acts as a weak acid. Since Cl- is the conjugate of a strong acid, HCl, that completely ionizes in solution, no significant amount of OH- will form, and the salt NH4Cl yields an acidic solution. When working with a polyprotic acid, at the first equivalence point, there will be just enough base to convert a polyprotic acid to its first salt. There will be an equivalence point as each proton is ionized in turn until only a salt with no ionizable protons remain. For example, the acid H₂A titrated with the base MOH would have a first equivalence at MHA, and a second at M₂ A. The pH of the salt where an ionizable proton remains, MHA, can be approximated using the equation pH at equivalence point with ionizable proton = PK₂1 +PK₂2 2 Part B What is the pH of a solution of 0.550 MK₂HPO4, potassium hydrogen phosphate? Express your answer numerically to the hundredths place. ► View Available Hint(s) 15. ΑΣΦ ? pH = Submit Part C What is the pH of a solution of 0.600 MKH₂PO4, potassium dihydrogen phosphate? Express your answer numerically to the hundredths place. ► View Available Hint(s) Π ΑΣΦ ? pH =