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9.4 pH and pOH

  • Page ID
    218433
  • If we wish to find the hydronium ion concentration ([H3O+]) and the pH of a solution, we need to know both the strength of the acid (or base) and the concentration of the acid (or base). We will find that we need to treat strong acids (and bases) differently than weak acids (and bases) based on the extent to which they react with water.

    Strong Acids

    By definition, strong acids are those acids with a \(K_a \geq 1\). Using this definition, we assume that strong acids will react completely with water, so that every molecule of acid reacts with a molecule of water to produce a hydronium ion and the conjugate base. (This assumption is, of course, not possible because there must always be at least a few particles of each reactant and each product at equilibrium, but our assumption is a valid one when \(K_a\) gets to be quite a bit larger than 1.):

    \[HA_{(aq)} \; + \; H_2O_{(l)} → H_3O^+_{(aq)} \; + \; A^-_{(aq)} \label {8.5.1} \]

    Based on our assumption, \([H_3O^+]_{eq} = [HA]_{initial} \), so we can carry out the following calculation:

    \[ pH = -log([H_3O^+]_{eq}) = -log([HA]_{initial}) \label {8.5.2}\]

    Example \(\PageIndex{1}\)

    What is the pH of a 0.175 M aqueous solution of HBr?

    Solution

    HBr is a strong acid (\(K_a > 1\)), so \([H_3O^+]_{eq} = [HBr]_{initial} \). Thus, pH = -log (0.175) = 0.757

    Exercise \(\PageIndex{1}\)

    What is the pH of a 0.0043 M aqueous solution of HNO3?

    Answer

    HNO3 is a strong acid (\(K_a > 1\)), so \([H_3O^+]_{eq} = [HNO_3]_{initial} \). Thus, pH = -log (0.0043) = 2.37

    Strong Bases

    By definition, strong bases are those bases with a \(K_b \geq 1\). using this definition, we assume that strong bases will react completely with water, so that every molecule of base reacts with a molecule of water to produce a hydroxide ion and the conjugate acid. (This assumption is, of course, not possible because there must always be at least a few particles of each reactant and each product at equilibrium, but our assumption is a valid one when \(K_b\) gets to be quite a bit larger than 1.):

    \[B_{(aq)} \; + \; H_2O_{(l)} → OH^-_{(aq)} \; + \; HB^+_{(aq)} \label {8.5.3} \]

    Based on our assumption, \([OH^-]_{eq} = [B]_{initial} \), so we can carry out the following calculation:

    \[ pOH = -log([OH^-]_{eq}) = -log([B]_{initial}) \label {8.5.4}\]

    To find the pH, we then subtract the pOH from 14.

    Example \(\PageIndex{2}\)

    What is the pH of a 0.175 M aqueous solution of NaNH2 ?

    Solution

    NH2- is a strong base (\(K_b > 1\)), so \([OH^-]_{eq} = [NH_2^- ]_{initial} \). Thus, pOH = -log (0.175) = 0.757, and pH = 14.000 - 0.757 = 13.243

    Exercise \(\PageIndex{2}\)

    What is the pH of a 0.0043 M aqueous solution of KOH?

    Answer

    KOH is a strong base (\(K_b = 1\)), so \([OH^-]_{eq} = [OH^-]_{initial} \). Thus, pOH = -log (0.0043) = 2.37, and pH = 14.00 - 2.37 = 11.63

    Weak Acids and Weak Bases

    If the acid or base you are working with is weak (\(K_a < 1 or K_b< 1\)), you must carry out amore complicated math problem to determine the \( [H_3O^+] \) because you cannot tell how much \(H_3O^+ \) will form in solution simply by looking at the intial \([HA] \). We will not learn these problem in this course.

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