In each case: a) Complete the acid-base equation with the help of tables if need
ID: 529328 • Letter: I
Question
In each case: a) Complete the acid-base equation with the help of tables if needed. b Label the two acids and bases involved. c) Explain whether the reactants or products are favoured at equilibrium. i. H_3PO_4 + C_2H_3O_2 D ii. C_2H_3O^-_2 + HPO_4 D iii. HPO^2-_4 + The^- D iv. SO^2-_3 + NH^+_4 D v. HO^-_2 + H_2CO_3 D One mL of 0.10 mol/L, Benzoic acid, C_6H_5COOH, is added to each of five test tubes containing 10.0 mL of 1.0 mol/L solutions of the following five ions. In each case write the acid-base equation. List the solutions in decreasing order of strength. The solutions: a. CO^2-_3 b. HCO^-_3 c. HPO^2-_4 d. CH_3COO^- e. HSO^-_3 List the following substances in decreasing order (strongest to weakest) of strength as acids and bases. Some may be used more than once. F H_2PO^-_4 H_2SO_4 HSO^-_3 HC_2O^-_4 NH_3, HSO^-_4Explanation / Answer
)
H3PO4 + C2H3O2-; note that H3PO4 is a stornger acid than acetic acid, so
H3PO4 = acid
C2H3O2- = base
C2H3O2H = conjugate acid
H2PO4- = conjugate base
H3PO4 + C2H3O2- --> H2PO4- + C2H3O2H
the products are strongly favoured
ii)
C2H3O2- + HPO4- --> C2H3O2H + PO4-3
C2H3O2- = base
HPO4- = acid
C2H3O2H = conjugate acid
PO4-3 = conjugate base
this is not favoured, since HPO4- is stronger base than C2H3O2- then, expect almost no reaction
iii)
HPO4-2 + HTe- <--> H2Te + PO4-
this is not favoured, since pKa1 = 11.4 and pKa = 11.3
this is almost in equilibrium
both species can act as either acid or bases, and form their respective acids, sicne they area amphoteric
iv
SO3-2 + NH4+ --> HSO3- + NH3
strongly favoured, since SO3-2 is a stornger base than NH4+
therefore, this favours products
SO3-2 = base
NH4+ = acid
HSO3- = conjugate acid
NH3 = conjguate base
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