Chapter 14: Q32 E (page 829)
Explain why the ionization constant, Ka, for HI is larger than the ionization constant for HF
Short Answer
HI has larger Ka as it is strong acid than HF.
Chapter 14: Q32 E (page 829)
Explain why the ionization constant, Ka, for HI is larger than the ionization constant for HF
HI has larger Ka as it is strong acid than HF.
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Get started for freeDetermine 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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