Question 10.10: What is the pH of a 0.0045 M solution of HClO4? ANALYSIS Fin...

What is the pH of a 0.0045 M solution of HClO_4?
ANALYSIS Finding pH requires first finding [H_3O^+] and then using the equation pH = – \log [H_3O^+]. Since HClO_4 is a strong acid (see Table 10.1), it is 100% dissociated, and so the H_3O^+ concentration is the same as the HClO_4 concentration.
BALLPARK ESTIMATE Because [H^+] = 4.5 \times 10^{-3}  M is close to midway between 1 \times 10^{-2}  M and 1 \times 10^{-3}  M, the pH must be close to the midway point between 2.0 and 3.0. (Unfortunately, because the logarithm scale is not linear, trying to estimate the midway point is not a simple process.)

 

CONJUGATE BASE ACID
Increasing base strength \left. \begin{matrix} ClO_4 ^- & Perchlorate ion \\ HSO_4 ^- & Hydrogen \\ & sulfate ion \\ I^- & Iodide ion \\ Br^- & Bromide ion \\ Cl^- & Chloride ion \\ NO_3 ^- & Nitrate ion \end{matrix} \right\} \pmb{Little or no reaction as bases} \pmb{Strong acids: 100\% dissociated} \begin{cases} Perchloric acid & HClO_4 \\ Sulfuric acid & H_2SO_4 \\ \\ Hydriodic acid & HI \\ Hydrobromic acid & HBr \\ Hydrochloric acid & HCl \\ Nitric acid & HNO_3 \end{cases}
\pmb{ H_2O } \pmb{ Water } \pmb{ Hydronium ion } \pmb{ H_3O^+ }
\left. \begin{matrix} SO_4 ^{2-} & Sulfate ion \\ \\ H_2PO_4 ^- & Dihydrogen \\ & phosphate ion \\ NO_2 ^- & Nitrite ion \\ F^- & Fluoride ion \\ CH_3COO^- & Acetate ion \end{matrix} \right\} \pmb{Very weak bases} \pmb{Weak acids} \begin{cases} Hydrogen & HSO_4 ^- \\ sulfate ion \\ Phosphoric acid & H_3PO_4 \\ \\ Nitrous acid & HNO_2 \\ Hydrofluoric acid & HF \\ Acetic acid & CH_3COOH \end{cases}
\left. \begin{matrix} HCO_3 ^- & Bicarbonate ion \\ HPO_4 ^{2-} & Hydrogen \\ & phosphate ion \\ NH_3 & Ammonia \\ CN^- & Cyanide ion \\ CO_3 ^{2- } & Carbonate ion \\ PO_4 ^{3-} & Phosphate ion \\ \end{matrix} \right\} \pmb{Weak bases} \pmb{Very weak acids} \begin{cases} Carbonic acid & H_2CO_3 \\ Dihydrogen & H_2PO_4 ^- \\ phosphate ion \\ Ammonium ion & NH_4 ^+ \\ Hydrocyanic acid & HCN \\ Bicarbonate ion & HCO_3 ^- \\ Hydrogen & HPO_4 ^{2-} \\ phosphate ion \end{cases}
\left. \pmb{ OH^- } \pmb{ Hydroxide ion } \right\} \pmb{ Strong base} \pmb{ Water } \pmb{ H_2O }
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[H_3O^+] = 0.0045  M = 4.5 \times 10^{-3}  M . Taking the negative logarithm gives pH = 2.35.
BALLPARK CHECK The calculated pH is consistent with our estimate.

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