One method is to apply a beneficial solvent such as anhydrous acetic acid

One method is to apply a beneficial solvent such as anhydrous acetic acid

Because acetic acid is a stronger acid than water, it must also be a weaker base, with a lesser tendency to accept a proton than \(H_2O\). Measurements of the conductivity of 0.1 M solutions of both HI and \(HNO_3\) in acetic acid show that HI is completely dissociated, but \(HNO_3\) is only partially dissociated and behaves like a weak acid in this solvent. This result clearly tells us that HI is a stronger acid than \(HNO_3\). The relative order of acid strengths and approximate \(K_a\) and \(pK_a\) values for the strong acids at the top of Table \(\PageIndex<1>\) were determined using measurements like this and different nonaqueous solvents.

From inside the aqueous selection, \(H_3O^+\) is the most powerful acid and you will \(OH^?\) ‘s the most powerful foot that can exist in equilibrium with \(H_2O\).

The leveling effect applies to solutions of strong bases as well: In aqueous solution, any base stronger than OH? is leveled to the strength of OH? because OH? is the strongest base that can exist in equilibrium with water. Salts such as \(K_2O\), \(NaOCH_3\) (sodium methoxide), and \(NaNH_2\) (sodamide, or sodium amide), whose anions are the conjugate bases of species that would lie below water in Table \(\PageIndex<2>\), are all strong bases that react essentially completely (and often violently) with water, accepting a proton to give a solution of \(OH^?\) and the corresponding cation:

Polyprotic Acids and you may Angles

Because you read, polyprotic acids eg \(H_2SO_4\), \(H_3PO_4\), and \(H_2CO_3\) contain sigbificantly more than simply you to ionizable proton, while the protons is actually forgotten from inside the a good stepwise manner. The new completely protonated types is always the most powerful acid because it is a lot easier to eliminate a beneficial proton from a simple molecule than just away from an effective adversely recharged ion. Therefore acidic strength decreases toward loss of further protons, and you can, correspondingly, the latest \(pK_a\) expands. Thought \(H_2SO_4\), such as for instance:

The hydrogen sulfate ion (\(HSO_4^?\)) is both the conjugate base of \(H_2SO_4\) and the conjugate acid of \(SO_4^<2?>\). Just like water, HSO4? can therefore act as either an acid or a base, depending on whether the other reactant is a stronger acid or a stronger base. Conversely, the sulfate ion (\(SO_4^<2?>\)) is a polyprotic base that is capable of accepting two protons in a stepwise manner:

In contrast, in the second reaction, appreciable quantities of both \(HSO_4^?\) and \(SO_4^<2?>\) are present at equilibrium

Like any other conjugate acidbase pair, the strengths of the conjugate acids and bases are related by \(pK_a\) + \(pK_b\) = pKw. Consider, for example, the \(HSO_4^?/ SO_4^<2?>\) conjugate acidbase pair. From Table \(\PageIndex<1>\), we see that the \(pK_a\) of \(HSO_4^?\) is 1.99. Hence the \(pK_b\) of \(SO_4^<2?>\) is ? 1.99 = . Thus sulfate is a rather weak base, whereas \(OH^?\) is a strong base, so the equilibrium shown in Equation \(\ref<16.6>\) lies to the left. The \(HSO_4^?\) ion is also a very weak base (\(pK_a\) of \(H_2SO_4\) = 2.0, \(pK_b\) of \(HSO_4^? = 14 ? (?2.0) = 16\)), which is consistent with what we expect for the conjugate base of a strong acid.

  • \(NH^+_<4(aq)>+PO^<3?>_ <4(aq)>\rightleftharpoons NH_<3(aq)>+HPO^<2?>_<4(aq)>\)
  • \(CH_3CH_2CO_2H_<(aq)>+CN^?_ <(aq)>\rightleftharpoons CH_3CH_2CO^?_<2(aq)>+HCN_<(aq)>\)

Identify the conjugate acidbase pairs in each reaction. Then refer to Tables \(\PageIndex<1>\)and\(\PageIndex<2>\) and Figure \(\PageIndex<2>\) to determine which is the stronger acid and base. Equilibrium always favors the formation of the weaker acidbase pair.

The conjugate acidbase pairs are \(NH_4^+/NH_3\) and \(HPO_4^<2?>/PO_4^<3?>\). According to Tables \(\PageIndex<1>\) and \(\PageIndex<2>\), \(NH_4^+\) is a stronger acid (\(pK_a = 9.25\)) than \(HPO_4^<2?>\) (pKa = ), and \(PO_4^<3?>\) is a stronger base (\(pK_b = 1.68\)) than \(NH_3\) (\(pK_b = 4.75\)). The equilibrium will therefore lie to the right, favoring the formation of the weaker acidbase pair:

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