t Salts of a Weak Polyprotic Acid salts form acidic or basic solutions through t
ID: 506635 • Letter: T
Question
t Salts of a Weak Polyprotic Acid 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 NHACI vill ionize in solution to form NH4+ and Cl The conjugate of NH3 will form some of the base through the equation and thus NH4 i 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 NH Cl 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 h the base MOH would have a first equivalence at MHA and a second at M2A. The pH of the salt where an ionizable proton remains, can be approximated using the equation pH at equivalence point with ionizable proton At 25 C phosphoric acid, H3PO4, has the following equilibrium constants: H3PO1 (aq)+ H2O(l Halo' (aq) +H2PO4 (aq) 7.5 x 10 H2PO4 (aq) +H2O H30 (aq)+HPO (aq) 6.2 x 10 HPO42 (aq)+ H20 (l H301 (aq) +PO13 (aq) Kala 4.2 x 10 Part A What is the pH of a solution of 0.25 MK3PO4, potassium phosphate? Express your answer numerically to the hundredths place. pH 2.82 Submit Hints My Answers Give Up View Part Correct You needed to calcluate the Kb for the reaction PO, (aq) +H2O) HPO, (aq) OH (aq) And then set up an equation that set Kb equal to IOH at equilibrium. Since the value of Kb is comparable in size to the joH you needed to solve a quadratic equation to find the concentration. Part BExplanation / Answer
K2HPO4 and KH2PO4 are amphoteric salts the pH values depends on pKa values but not formal concentraions
part B )
pH = (pKa2 + pKa3) / 2
pH = (7.21 + 12.38) / 2
pH = 9.79
part C )
pH = (pKa1 + pKa2) / 2
pH = (2.12 +7.21) / 2
pH = 4.67
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