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Chapter 14: Q14.3-59 E (page 832)

Determine \({K_a}\)for hydrogen sulfate ion, \(HS{O_4}^ - .\)In a \(0.10 - M\)solution the acid is \(29\% \)ionized.

Short Answer

Expert verified

The acid ionization constant for \({\rm{HSO}}_4^ - \) is \(0.01\).

Step by step solution

01

Required formulas to solve

  • The reaction in question is \(HSO_4^ - (aq) \to {H^ + }(aq) + SO_4^{2 - }(aq).\)
  • Now, the \({K_a}\) value can be calculate as: \({K_a} = \frac{{c\left( {{H^ + }} \right) \cdot c\left( {SO_4^{2 - }} \right)}}{{c\left( {{\rm{HSO}}_4^ - } \right)}}.\)
  • The percentage of the ionization (alpha) can be calculated as: \(\alpha = \frac{{c\left( {{H^ + }} \right)}}{{{c_0}\left( {HSO_4^ - } \right)}}.\)
02

Determine the acid ionization constant

  • As \({c_0}\)is the starting concentration, with these two values, let us calculate the equilibrium concentrations of \({{\rm{H}}^ + }\)and \({\rm{SO}}_4^{2 - }.\)
  • Let us assume that they are the same from the reaction equation.
    \(\begin{aligned}{\rm{SO}}_4^{2 - }c\left( {{H^ + }} \right) &= 29\% \cdot 0.10M\\c\left( {{H^ + }} \right) &= 0.03M.\end{aligned}\)
  • Further, let the equilibrium concentration of \({\rm{HSO}}_4^ - \) be x.If we mark the concentration change from the initial point until the equilibrium is reached as\(x,\) then the concentration of \({\rm{HSO}}_4^ - \)drops by \(x\) and the concentration of \({H^ + }\)and \({\rm{SO}}_4^{2 - }\)rises by \(x\).
  • If we assume that the initial concentrations of \({{\rm{H}}^ + }\)and \(S{\rm{O}}_4^{2 - }\)is \(0{\rm{M}}\), the change equals \(x = 0.03M.\)
  • Therefore, the equilibrium concentration of \({\rm{HSO}}_4^ - \) is:
    \(\begin{aligned}c\left( {HSO_4^ - } \right) &= {c_0}\left( {SO_4^{2 - }} \right) - x\\c\left( {HSO_4^ - } \right) &= (0.10 - 0.03)M\\c\left( {HSO_4^ - } \right) &= 0.07M.\end{aligned}\)
  • Now, let us calculate the \({K_a}:\)
    \(\begin{aligned}{K_a} &= \frac{{0.03M \cdot 0.03M}}{{0.07M}}\\{K_a} &= 0.01.\end{aligned}\)
  • Hence, the acid ionization constant for \({\rm{HSO}}_4^ - \) is \(0.01\).

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Most popular questions from this chapter

From the equilibrium concentrations given, calculate for each of the weak acids and for each of the weak bases.

\(\begin{aligned}(a)C{H_3}C{O_2}H:\left( {{H_3}{O^ + }} \right) = 1.34 \times 1{0^{ - 3}}M;\left( {C{H_3}CO_2^ - } \right) = 1.34 \times 1{0^{ - 3}}M;\left( {C{H_3}C{O_2}H} \right) = 9.866 \times 1{0^{ - 2}}M;\\(b)Cl{O^ - }:\left( {O{H^ - }} \right) = 4.0 \times 1{0^{ - 4}}M;(HClO) = 2.38 \times 1{0^{ - 5}}M;\left( {Cl{O^ - }} \right) = 0.273M;\\(c)HC{O_2}H:\left( {HC{O_2}H} \right) = 0.524M;\left( {{H_3}{O^ + }} \right) = 9.8 \times 1{0^{ - 3}}M\left( {HCO_2^ - } \right) = 9.8 \times 1{0^{ - 3}}M;\\(d){C_6}{H_5}NH_3^ + :\left( {{C_6}{H_5}NH_3^ + } \right) = 0.233M;\left( {{C_6}{H_5}N{H_2}} \right) = 2.3 \times 1{0^{ - 3}}M;\left( {{H_3}{O^ + }} \right) = 2.3 \times 1{0^{ - 3}}M\end{aligned}\)

Nicotine, \({C_{10}}{H_{14}}\;{N_2}\), is a base that will accept two protons \(\left( {{K_1} = 7 \times 1{0^{ - 7}},{K_2} = 1.4 \times 1{0^{ - 11}}} \right)\). What is the concentration of each species present in a \(0.050 - M\) solution of nicotine?

Calculate \(pH\;and the\;pOH\) of each of the following solutions at\(2{5^o}C\)for which the substances ionize completely:

(a)\(0.200M HCl\)

(b)\(0.0143M NaOH\)

(c)\(3.0M HN{O_3}\)

(d) \(0.0031M Ca{(OH)_2}\)

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\(\left( {{K_w}} \right)\;is\;9.311 \times 1{0^{ - 14}}\;at\;6{0^o}C\). Calculate \(\left( {{H_3}{O^ + }} \right),\left( {O{H^ - }} \right),pH,\;and\;pOH\)for pure water at \(6{0^o}C\).
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