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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

Saccharin, \({{\bf{C}}_{\bf{7}}}{{\bf{H}}_{\bf{4}}}{\bf{NS}}{{\bf{O}}_{\bf{3}}}{\bf{H}}\), is a weak acid\(\left( {{\bf{Ka = 2}}{\bf{.1 \times 1}}{{\bf{0}}^{{\bf{ - 2}}}}} \right)\). If \({\bf{0}}.{\bf{250}}{\rm{ }}{\bf{L}}\) of diet cola with a buffered pH of \({\bf{5}}.{\bf{48}}\) was prepared from \({\bf{2}}.{\bf{00}}{\rm{ }} \times {\rm{ }}{\bf{1}}{{\bf{0}}^{ - {\bf{3}}}}{\bf{g}}\) of sodium saccharide, \({\bf{Na}}\left( {{{\bf{C}}_{\bf{7}}}{{\bf{H}}_{\bf{4}}}{\bf{NS}}{{\bf{O}}_{\bf{3}}}} \right)\), what are the final concentrations of saccharine and sodium saccharide in the solution?

Propionic acid, \({C_2}{H_5}C{O_2}H\left( {{K_a} = 1.34 \times 1{0^{ - 5}}} \right)\), is used in the manufacture of calcium propionate, a food preservative. What is the hydronium ion concentration in a \(0.698 M\) solution of \({C_2}{H_5}C{O_2}H ? \)

Is the self ionization of water endothermic or exothermic? The ionization constant for water \(\left( {{K_W}} \right)\)is 2.9* \({10^{ - 14}}\)at \({40^ \circ }{\rm{C}}\)and 9.3 x \({10^{ - 14}}\)at \({60^ \circ }{\rm{C}}\)

Both \(HF and HCN\)ionize in water to a limited extent. Which of the conjugate bases \(F - or CN - \), is the stronger base? See Table 14.3.

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

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

(b) \( RbI\)

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