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Chapter 14: Q14.2-17 E (page 828)

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 \(6{0^o}C\).

Short Answer

Expert verified

The value of \(\left( {{{\rm{H}}_3}{{\rm{O}}^ + }} \right)\) for pure water at\({60^ \circ }C\)is\(c\left( {{H_3}{O^ + }} \right) = 3.05 \cdot {10^{ - 7}}\frac{{mol}}{L}\).

The value of \(\left( {{\rm{O}}{{\rm{H}}^ - }} \right)\)for pure water at\({60^ \circ }C\)is\(c\left( {O{H^ - }} \right) = 3.05 \cdot {10^{ - 7}}\frac{{mol}}{L}\).

The \(pH\) for pure water at\({60^ \circ }C\)is\(pH = 6.51\)

The \(pOH\)for pure water at\({60^ \circ }C\)is\(pOH = 6.51\)

Step by step solution

01

Define the formula for\(pH\)and\(pOH\):

The formula for\(pH\)is\(pH = - \log \frac{{c\left( {{H_3}{O^ + }} \right)}}{{mol{L^{ - 1}}}}\).

The formula for\(pOH\)is\(pOH = - \log \frac{{c\left( {O{H^ - }} \right)}}{{mol{L^{ - 1}}}}\).

02

Obtain the solution by using the formula:

Let the ionization constant for water\(\left( {{K_w}} \right){\rm{\;is\;}}9.311 \times {10^{ - 14}}{\rm{\;at\;}}{60^ \circ }{\rm{C}}\).

Thus, the equation of water ionization is,

\({{\rm{H}}_2}{\rm{O}}(l) + {{\rm{H}}_2}{\rm{O}}(l) \to {{\rm{H}}_3}{{\rm{O}}^ + }(aq) + {\rm{O}}{{\rm{H}}^ - }(aq)\)

Hence, the ionization constant is,

\({K_w} = c\left( {{H_3}{O^ + }} \right) \cdot c\left( {O{H^ - }} \right)\)

Since, the concentration of\({{\rm{H}}_3}{{\rm{O}}^ + }{\rm{and\;O}}{{\rm{H}}^ - }\)are equal, it can be written as,

\(c\left( {{H_3}{O^ + }} \right) = c\left( {O{H^ - }} \right) = c\)

Find the concentration,

\(\begin{aligned}{9.311 \cdot {{10}^{ - 14}} = {c^2}}\\{c = \sqrt {9.311 \cdot {{10}^{ - 14}}} }\\{c = 3.05 \cdot {{10}^{ - 7}}\frac{{{\rm{mol}}}}{{\rm{L}}}}\end{aligned}\)

Therefore,

\(\begin{aligned}{c\left( {{H_3}{O^ + }} \right) = 3.05 \cdot {{10}^{ - 7}}\frac{{mol}}{L}}\\{c\left( {O{H^ - }} \right) = 3.05 \cdot {{10}^{ - 7}}\frac{{mol}}{L}}\end{aligned}\)

03

Calculate the value of \(pH\)and \(pOH\):

Use the formula,

\(pH = - \log \frac{{c\left( {{H_3}{O^ + }} \right)}}{{mol{L^{ - 1}}}}\)

Substitute, \(c\left( {{H_3}{O^ + }} \right) = 3.05 \cdot {10^{ - 7}}\frac{{mol}}{L}\)

\(\begin{aligned}{pH = - \log \frac{{3.05 \cdot {{10}^{ - 7}}{\rm{mol}}{{\rm{L}}^{ - 1}}}}{{{\rm{mol}}{L^{ - 1}}}}}\\{pH = 6.51}\end{aligned}\)

Calculate \(pOH,\)

\(pOH = - \log \frac{{c\left( {O{H^ - }} \right)}}{{mol{L^{ - 1}}}}\)

Substitute, \(c\left( {O{H^ - }} \right) = 3.05 \cdot {10^{ - 7}}\frac{{mol}}{L}\)

\(pOH = - \log \frac{{3.05 \cdot {{10}^{ - 7}}mol{L^{ - 1}}}}{{mol{L^{ - 1}}}}\)

\(pOH = 6.51\)

Therefore, the value of\(pH\)is\(6.51\)and the value of\(pOH\)is \(6.51.\)

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Most popular questions from this chapter

We can ignore the contribution of water to the concentration of\(O{H^ - }\)in a solution of the following bases:\(0.0784 M {C_6}{H_5}N{H_2}\), a weak base\(0.11M{\left( {C{H_3}} \right)_3}\;N\), a weak base but not the contribution of water to the concentration of\({H_3}{O^ + }?\)

Which of the following will increase the percentage of \({\rm{N}}{{\rm{H}}_3}\)that is converted to the ammonium ion in water (Hint: Use Le Châtelier's principle)?

(a) Addition of \(NaOH\)

(b) Addition of \(HCl\)

(c) Addition of \(N{H_4}Cl\)

A buffer solution is prepared from equal volumes of \({\bf{0}}.{\bf{200}}{\rm{ }}{\bf{M}}\) acetic acid and \({\bf{0}}.{\bf{600}}{\rm{ }}{\bf{M}}\) sodium acetate. Use \({\bf{1}}.{\bf{80}}{\rm{ }} \times {\rm{ }}{\bf{1}}{{\bf{0}}^{ - {\bf{5}}}}\) as Ka for acetic acid.

(a) What is the pH of the solution?

(b) Is the solution acidic or basic?

(c) What is the pH of a solution that results when \({\bf{3}}.{\bf{00}}{\rm{ }}{\bf{mL}}\) of \({\bf{0}}.{\bf{034}}{\rm{ }}{\bf{M}}{\rm{ }}{\bf{HCl}}\) is added to \({\bf{0}}.{\bf{200}}{\rm{ }}{\bf{L}}\) of the original buffer?

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

(a)\(0.000259M HCl{O_4}\)

(b)\(0.21M NaOH\)

(c)\(0.000071M Ba{(OH)_2}\)

(d) \(2.5M KOH\)

Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:

\({\rm{\;(a)\;NO}}_2^ - + {{\rm{H}}_2}{\rm{O}} \to {\rm{HN}}{{\rm{O}}_2} + {\rm{O}}{{\rm{H}}^ - }\).

\({\rm{\;(b)\;HBr}} + {{\rm{H}}_2}{\rm{O}} \to {{\rm{H}}_3}{{\rm{O}}^ + } + {\rm{B}}{{\rm{r}}^ - }\)

\({\rm{\;(c)\;H}}{{\rm{S}}^ - } + {{\rm{H}}_2}{\rm{O}} \to {{\rm{H}}_2}{\rm{S}} + {\rm{O}}{{\rm{H}}^ - }\)

\({\rm{\;(d)\;}}{{\rm{H}}_2}{\rm{PO}}_4^ - + {\rm{O}}{{\rm{H}}^ - } \to {\rm{HP}}{{\rm{O}}_4}^{2 - } + {{\rm{H}}_2}{\rm{O}}\)

\({\rm{\;(e)\;}}{{\rm{H}}_2}{\rm{PO}}_4^ - + {\rm{HCl}} \to {{\rm{H}}_3}{\rm{P}}{{\rm{O}}_4} + {\rm{C}}{{\rm{l}}^ - }\)

\({\rm{\;(f)\;}}{\left( {{\rm{Fe}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_5}({\rm{OH}})} \right)^{2 + }} + {\left( {{\rm{Al}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_6}} \right)^{3 + }} \to {\left( {{\rm{Fe}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_6}} \right)^{3 + }} + {\left( {{\rm{Al}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_5}({\rm{OH}})} \right)^{2 + }}\)

\({\rm{\;(g)\;C}}{{\rm{H}}_3}{\rm{OH}} + {{\rm{H}}^ - } \to {\rm{C}}{{\rm{H}}_3}{{\rm{O}}^ - } + {{\rm{H}}_2}\)
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