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14.8: Water: Acid and Base in One

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  • Page ID
    98105
  • Skills to Develop

    • Describe the autoionization of water.
    • Calculate the concentrations of \(\ce{H3O^{+}}\) and \(\ce{OH^{−}}\) in aqueous solutions, knowing the other concentration.

    We have already seen that \(\ce{H2O}\) can act as an acid or a base:

    \[\color{blue}{\underbrace{\ce{NH3}}_{\text{base}}} + \color{red}{\underbrace{\ce{H2O}}_{\text{acid}}} \color{black} \ce{<=> NH4^{+} + OH^{−}}\]

    where \(\ce{H2O}\) acts as an \(\color{red}{\text{acid}}\) (in red).

    \[\color{red}{\underbrace{\ce{HCl}}_{\text{acid}}} + \color{blue}{\underbrace{\ce{H2O}}_{\text{base}}} \color{black} \ce{-> H3O^{+} + Cl^{−}}\]

    where \(\ce{H2O}\) acts as an \(\color{blue}{\text{base}}\) (in blue).

    It may not surprise you to learn, then, that within any given sample of water, some \(\ce{H2O}\) molecules are acting as acids, and other \(\ce{H2O}\) molecules are acting as bases. The chemical equation is as follows:

    \[\color{red}{\underbrace{\ce{H2O}}_{\text{acid}}} + \color{blue}{\underbrace{\ce{H2O}}_{\text{base}}} \color{black} \ce{<=> H3O^{+} + OH^{−}} \label{Auto}\]

    This occurs only to a very small degree: only about 6 in 108 \(\ce{H2O}\) molecules are participating in this process, which is called the autoionization of water.

    Figure \(\PageIndex{1}\): Autoionization of water, giving hydroxide and hydronium ions.

    At this level, the concentration of both \(\ce{H3O^{+}(aq)}\) and \(\ce{OH^{−}(aq)}\) in a sample of pure \(\ce{H2O}\) is about \(1.0 \times 10^{−7}\, M\) (at room temperature). If we use square brackets—[ ]—around a dissolved species to imply the molar concentration of that species, we have

    \[\color{red}{\ce{[H3O^{+}]}} \color{black}{ = } \color{blue}{\ce{[OH^{-}]}} \color{black} = 1.0 \times 10^{-7} \label{eq5}\]

    for any sample of pure water because H2O can act as both an acid and a base. The product of these two concentrations is \(1.0\times 10^{−14}\):

    \[\color{red}{\ce{[H3O^{+}]}} \color{black}{\times} \color{blue}{\ce{[OH^{-}]}} \color{black} = (1.0 \times 10^{-7})( 1.0 \times 10^{-7}) = 1.0 \times 10^{-14}\]

    • For acids, the concentration of \(\ce{H3O^{+}(aq)}\) (i.e., \(\ce{[H3O^{+}]}\)) is greater than \(1.0 \times 10^{−7}\, M\).
    • For bases the concentration of \(\ce{OH^{−}(aq)}\) (i.e., \(\ce{[OH^{−}]}\)) is greater than \(1.0 \times 10^{−7}\, M\).

    However, the product of the two concentrations—\(\ce{[H3O^{+}][OH^{−}]}\)—is always equal to \(1.0 \times 10^{−14}\), no matter whether the aqueous solution is an acid, a base, or neutral:

    \[\color{red}{\ce{[H_3O^+]}} \color{blue}{\ce{[OH^{-}]}} \color{black} = 1.0 \times 10^{-14} \]

    This value of the product of concentrations is so important for aqueous solutions that it is called the autoionization constant of water and is denoted \(K_w\):

    \[K_w = \color{red}{\ce{[H_3O^+]}} \color{blue}{\ce{[OH^{-}]}} \color{black} = 1.0 \times 10^{-14} \label{eq10}\]

    This means that if you know \(\ce{[H3O^{+}]}\) for a solution, you can calculate what \(\ce{[OH^{−}]}\)) has to be for the product to equal \(1.0 \times 10^{−14}\), or if you know \(\ce{[OH^{−}]}\)), you can calculate \(\ce{[H3O^{+}]}\). This also implies that as one concentration goes up, the other must go down to compensate so that their product always equals the value of \(K_w\).

    Warning: Temperature Matters

    The degree of autoionization of water (Equation \ref{Auto}) and hence the value of \(K_w\) changes with temperature, so Equations \ref{eq5} - \ref{eq10} are accurate only at room temperature.

    Example \(\PageIndex{1}\): Hydroxide ConCentration

    What is \(\ce{[OH^{−}]}\)) of an aqueous solution if \(\ce{[H3O^{+}]}\) is \(1.0 \times 10^{−4} M\)?

    SOLUTION

    Steps for Problem Solving  
    Identify the "given" information and what the problem is asking you to "find."

    Given: \(\ce{[H3O^{+}]}  =1.0 \times 10^{−4}\, M\)

    Find:  [OH] = ? M

    List other known quantities

    none

    Plan the problem

    Using the expression for \(K_w\), (Equation \ref{eq10}), rearrange the equation algebraically to solve for [OH].

    \[\left [ \ce{OH^{-}} \right ]=\dfrac{1.0\times 10^{-14}}{\left [ H_3O^+ \right ]} \nonumber\]

    Calculate 

    Now substitute the known quantities into the equation and solve.

    \[\left [\ce{ OH^{-}} \right ]=\dfrac{1.0\times 10^{-14}}{1.0\times 10^{-4}}=1.0\times 10^{-10}M\nonumber\]

    It is assumed that the concentration unit is molarity, so \(\ce{[OH^{−}]}\) is 1.0 × 10−10 M.

    Think about your result. The concentration of the acid is high (> 1 x 10-7 M) so \(\ce{[OH^{−}]}\) should be low.

    Exercise \(\PageIndex{1}\)

    What is \(\ce{[OH^{−}]}\) in a 0.00032 M solution of H2SO4?

    Hint

    Assume both protons ionize from the molecule.. although this is not the case.

    Answer
    \(3.1 \times 10^{−11}\, M\)

    When you have a solution of a particular acid or base, you need to look at the formula of the acid or base to determine the number of H3O+ or OH ions in the formula unit because \(\ce{[H_3O^{+}]}\) or \(\ce{[OH^{−}]}\)) may not be the same as the concentration of the acid or base itself.

    Example \(\PageIndex{2}\): Hydronium Concentration

    What is \(\ce{[H_3O^{+}]}\) in a 0.0044 M solution of \(\ce{Ca(OH)_2}\)?

    SOLUTION

    Steps for Problem Solving

     
    Identify the "given"information and what the problem is asking you to "find."

    Given: \([\ce{Ca(OH)_2}] =0.0044 \,M\)

     Find: \(\ce{[H_3O^{+}]}\) = ? M

    List other known quantities

    We begin by determining \(\ce{[OH^{−}]}\). The concentration of the solute is 0.0044 M, but because \(\ce{Ca(OH)_2}\) is a strong base, there are two OH ions in solution for every formula unit dissolved, so the actual \(\ce{[OH^{−}]}\) is two times this,

    \[\ce{[OH^{−}] = 2 \times 0.0044\, M = 0.0088 \,M.} \nonumber\]

    Plan the problem

    Use the expression for \(K_w\) (Equation \ref{eq10}) and rearrange the equation algebraically to solve for \(\ce{[H_3O^{+}]}\).

    \[\left [ H_3O^{+} \right ]=\dfrac{1.0\times 10^{-14}}{\left [ OH^{-} \right ]} \nonumber\]

    Calculate 

    Now substitute the known quantities into the equation and solve.

    \[\left [ H_3O^{+} \right ]=\frac{1.0\times 10^{-14}}{(0.0088)}=1.1\times 10^{-12}M \nonumber\]

    \(\ce{[H_3O^{+}]}\) has decreased significantly in this basic solution.

    Think about your result. The concentration of the base is high (> 1 x 10-7 M) so \(\ce{[H_3O^+}]}\) should be low.

    Exercise \(\PageIndex{2}\)

    What is \(\ce{[H_3O^{+}]}\) of an aqueous solution if \(\ce{[OH^{−}]}\) is \(1.0 \times 10^{−9}\, M\)?

    Answer
    1.0 × 10−5

    In any aqueous solution, the product of \(\ce{[H_3O^{+}]}\) and \(\ce{[OH^{−}]}\) equals \(1.0 \times 10^{−14}\) (at room temperature).

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