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Chapter 13: Q13E (page 750)

For a titration to be effective, the reaction must be rapid and the yield of the reaction must essentially be 100%.

Is \({K_c} > 1,\; < 1\), or \( \approx 1\) for a titration reaction?

Short Answer

Expert verified

As titration reactions are rapid and yield is almost 100%, so \({K_c} > 1.\)

Step by step solution

01

Definition of Titration reactions

The term titration reaction refers to a reaction in which one substance is entirely neutralized by another substance. This titration could be an acid-base titration, a redox titration, or something else entirely.

02

Find the reaction yield

Let's consider an acid base titration reaction as

\(\begin{aligned}{}{\rm{HCl}}(aq) + {\rm{NaOH}}(aq) \to {\rm{NaCl}}(aq) + {{\rm{H}}_2}{\rm{O}}(aq)\\{\rm{HCl}}(aq) + {\rm{NaOH}}(aq) \to {\rm{NaCl}}(aq) + {{\rm{H}}_2}{\rm{O}}(aq)\end{aligned}\)

This reaction has a 100% yield, which means that all of the starting materials have been transformed to products and there are no leftovers. So,\({K_c} > 1\).

Thus,\({K_c} > 1\)for all the titration reactions which have 100% yield.

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Most popular questions from this chapter

The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.

(a) \(\begin{aligned}{}2{\rm{N}}{{\rm{H}}_3}(\;{\rm{g}}) \rightleftharpoons {{\rm{N}}_2}(\;{\rm{g}}) + 3{{\rm{H}}_2}(\;{\rm{g}})\;\;\;{K_e} b= 17;\\\;\;\;\left( {{\rm{N}}{{\rm{H}}_3}} \right) = 0.50\,M,\;\left( {{{\rm{N}}_2}} \right) = 0.15\,M,\;\left( {{{\rm{H}}_2}} \right) = 0.12\,M\end{aligned}\)

(b) \(\begin{aligned}{}2{\rm{N}}{{\rm{H}}_3}(\;g) \rightleftharpoons {{\rm{N}}_2}(\;g) + 3{{\rm{H}}_2}(\;g)\;\;\;{K_P} = 6.8 \times {10^4};\\\;\;\;\;{\rm{initial}}\;{\rm{pressures:}}\,{\kern 1pt} {\rm{N}}{{\rm{H}}_3} = 2.00\;{\rm{atm}},\;{\kern 1pt} {\kern 1pt} {{\rm{N}}_2} = 10.00\,{\rm{atm}},\;{{\rm{H}}_2} = 10.00\,{\rm{atm}}\end{aligned}\)

(c) \(\begin{aligned}{}2{\rm{S}}{{\rm{O}}_3}(\;g) \rightleftharpoons 2{\rm{S}}{{\rm{O}}_2}(\;g) + {{\rm{O}}_2}(\;g)\;\;\;{K_c} = 0.230;\\\;{\kern 1pt} {\kern 1pt} {\kern 1pt} \;\left( {{\rm{S}}{{\rm{O}}_3}} \right) = 2.00\,M,\;\left( {{\rm{S}}{{\rm{O}}_2}} \right) = 2.00\,M,\;\left( {{{\rm{O}}_2}} \right) = 2.00\,M\end{aligned}\)

(d) \(\begin{aligned}{}2{\rm{S}}{{\rm{O}}_3}(\;g) \rightleftharpoons 2{\rm{S}}{{\rm{O}}_2}(\;g) + {{\rm{O}}_2}(\;g)\;\;\;{K_p} = 6.5\;{\rm{atm}};\\\;\;\;\;{\rm{initial}}\;{\rm{pressures:}}\;{\rm{S}}{{\rm{O}}_2} = 1.00\;{\rm{atm}},\;{{\rm{O}}_2} = 1.130\;{\rm{atm}},\;{\rm{S}}{{\rm{O}}_3} = 0\;{\rm{atm}}\end{aligned}\)

(e) \(\begin{aligned}{}2{\rm{NO}}(g) + {\rm{C}}{{\rm{l}}_2}(g) \rightleftharpoons 2{\rm{NOCl}}(g)\;\;\;{K_P} = 2.5 \times {10^3};\\\;\;\;\;{\rm{initial}}\;{\rm{pressures:}}\;{\rm{NO}} = 1.00\;{\rm{atm}},{\rm{C}}{{\rm{l}}_2} = 1.00\;{\rm{atm}},\;{\rm{NOCl}} = 0\;{\rm{atm}}\end{aligned}\)

(f) \(\begin{aligned}{}{{\rm{N}}_2}(\;g) + {{\rm{O}}_2}(\;g) \rightleftharpoons 2{\rm{NO}}(g)\;\;\;{K_c} = 0.050;\\\;\;\;\;\;\left( {{{\rm{N}}_2}} \right) = 0.100M,\;\left( {{{\rm{O}}_2}} \right) = 0.200M,\;({\rm{NO}}) = 1.00M\end{aligned}\)

Question: Consider the reaction between \({{\rm{H}}_2}\)and \({{\rm{O}}_2}\)at 100 K\({K_P} = \frac{{{{\left( {{P_{{{\rm{H}}_2}{\rm{O}}}}} \right)}^2}}}{{\left( {{P_{{{\rm{O}}_2}}}} \right){{\left( {{P_{{{\rm{H}}_2}}}} \right)}^2}}} = 1.33 \times {10^{20}}\)

If 0.500 atm of H2 and 0.500 atm of O2are allowed to come to equilibrium at this temperature, what are the partial pressures of the components?

Acetic acid is a weak acid that reacts with water according to this equation:

\(C{H_3}C{O_2}H(aq) + {H_2}O(aq) \rightleftharpoons {H_3}{O^ + }(aq) + C{H_3}CO_2^ - (aq)\)

Will any of the following increase the percent of acetic acid that reacts and produces \(C{H_3}CO_2^ - \)ion?

(a) Addition of \(HCl\)

(b) Addition of \(NaOH\)

(c) Addition of \(NaC{H_3}C{O_2}\)

Explain how to recognize the conditions under which changes in pressure would affect systems at equilibrium.

Calculate the equilibrium concentrations of NO, O2, and NO2 in a mixture at 250 掳C that results from the reaction of 0.20 M NO and 0.10 M O2. (Hint: K is large; assume the reaction goes to completion then comes back to equilibrium.)

\(2NO(g) + {O_2}(g) \rightleftharpoons 2N{O_2}(g)\quad {K_c} = 2.3 \times 1{0^5}\;at\;25{0^o}C\)
See all solutions

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