Explain why the ionization constant, Ka, for HI is larger than the ionization constant for HF

Short Answer

Expert verified

HI has larger Ka as it is strong acid than HF.

Step by step solution

01

Definition of the ionization constant 

Ka (acidity constant or acid-ionization constant) is a quantitative measure of an acid's strength in solution. In the context of acid-base processes, it is the equilibrium constant for a chemical reaction known as dissociation.

02

Calculating the dissociation constant of an acid

Ka is the constant of dissociation of acid as

\begin{aligned}{HA(aq)\to{H^+}(aq)+{A^-}(aq)}\\{{K_a}=\frac{{c\left({{H^+}}\right)\times c\left({{A^-}}\right)}}{{c(HA)}}}\end{aligned}

For stronger acids,Ka has greater dissociation. A strong acid has a stable conjugated base and does not react with conjugated acid to produce the reactant. A weak acid does not have dissociated conjugate base, is not stable, and reacts with conjugated acid to produce the reactant. Therefore, from the same initial concentration of weak and strong acids, strong acids have more protons. pH is lower in strong acids and higher in weak acids.

HI has largerKa, it is a stronger acid than HF×I- is a more stable conjugated base than F-, as it has a larger atomic radius.

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Most popular questions from this chapter

Determine whether aqueous solutions of the following salts are acidic, basic, or neutral

(a) \( Al{\left( {N{O_3}} \right)_3}\)

(b) \( RbI\)

(c) \( KHC{O_2}\)

(d) \( C{H_3}N{H_3}Br\)

The ionization constant for water\(({K_w})\) is\(2.9 \times 1{0^{ - 14}}\;at\;4{0^o}C\). Calculate\(\left( {{H_3}{O^ + }} \right),\left( {O{H^ - }} \right),pH,\;and\;pOH\) for pure water at \(4{0^o}C\).

What is the effect on the concentration of ammonia, hydroxide ion, and ammonium ion when the following are added to a basic buffer solution of equal concentrations of ammonia and ammonium nitrate:

(a)\(KI\)

(b)\(N{H_3}\)

(c)\(HI\)

(d)\(NaOH\)

(e)\(N{H_4}Cl\)

Predict which acid in each of the following pairs is the stronger and explain your reasoning for each. \(\;(a)\;{H_2}O\;or\;HF,\;(b)\;B{(OH)_3}\;or\;Al{(OH)_3},\;(c)\;HSO_3^ - or\;HSO_4^ - ,\;(d)\;N{H_3}\;or\;{H_2}S,\;(e)\;{H_2}O\;or\;{H_2}Te\)

From the equilibrium concentrations given, calculate \({K_a}\)for each of the weak acids and \({K_b}\)for each of the weak bases.

\(\begin{aligned}(a)N{H_3}:\left( {O{H^ - }} \right) = 3.1 \times 1{0^{ - 3}}M\left( {NH_4^ + } \right) = 3.1 \times 1{0^{ - 3}}M;\left( {N{H_3}} \right) = 0.533M;\\(b)HN{O_2}:\left( {{H_3}{O^ + }} \right) = 0.011M;\left( {NO_2^ - } \right) = 0.0438M;\left( {HN{O_2}} \right) = 1.07M;\\(c){\left( {C{H_3}} \right)_3}\;N:\left( {{{\left( {C{H_3}} \right)}_3}\;N} \right) = 0.25M;\left( {{{\left( {C{H_3}} \right)}_3}N{H^ + }} \right) = 4.3 \times 1{0^{ - 3}}M;\left( {O{H^ - }} \right) = 4.3 \times 1{0^{ - 3}}M;\\(d)N{H_4} + :\left( {N{H_4} + } \right) = 0.100M;\left( {N{H_3}} \right) = 7.5 \times 1{0^{ - 6}}M;\left( {{H_3}{O^ + }} \right) = 7.5 \times 1{0^{ - 6}}M\end{aligned}\)

\(\begin{aligned}\left( {N{H_3}} \right) = 7.5 \times 1{0^{ - 6}}M\\\left( {{H_3}{O^ + }} \right) = 7.5 \times 1{0^{ - 6}}M\end{aligned}\)

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