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Chapter 15: Problem 44

Calculate the \(\mathrm{pH}\) and \(\mathrm{pOH}\) of the following solutions. (a) \(0.51 \mathrm{M} \mathrm{CsOH}\) (b) \(0.0040 \mathrm{M} \mathrm{HI}\) (c) \(0.13 M \mathrm{LiOH}\) (d) \(0.66 \mathrm{MHClO}_{4}\)

Short Answer

Expert verified
(a) pH = 13.71, pOH = 0.29. (b) pH = 2.40, pOH = 11.60. (c) pH = 13.11, pOH = 0.89. (d) pH = 0.18, pOH = 13.82.

Step by step solution

01

Determine the type of solution

Identify whether each compound is a strong acid or a strong base. (a) CsOH is a strong base. (b) HI is a strong acid. (c) LiOH is a strong base. (d) HClO鈧� is a strong acid.
02

Calculating pH for strong bases

For strong bases, calculate the concentration of hydroxide ions, \([\text{OH}^-]\), and find the \(\text{pOH}\) first:(a) CsOH dissociates completely to give [OH鈦籡 = 0.51 M. \(\text{pOH} = -\log([\text{OH}^-]) = -\log(0.51) = 0.292\).(c) LiOH dissociates completely to give [OH鈦籡 = 0.13 M. \(\text{pOH} = -\log([\text{OH}^-]) = -\log(0.13) = 0.886\).Then calculate \(\text{pH}\) using \(\text{pH} = 14 - \text{pOH}\):(a) \(\text{pH} = 14 - 0.292 = 13.708\).(c) \(\text{pH} = 14 - 0.886 = 13.114\).
03

Calculating pH for strong acids

For strong acids, calculate the concentration of hydrogen ions, \([\text{H}^+]\), and find \(\text{pH}\):(b) HI dissociates completely to give [H鈦篯 = 0.0040 M. \(\text{pH} = -\log([\text{H}^+]) = -\log(0.0040) = 2.398\).(d) HClO鈧� dissociates completely to give [H鈦篯 = 0.66 M. \(\text{pH} = -\log([\text{H}^+]) = -\log(0.66) = 0.18\).
04

Calculating pOH for strong acids

For strong acids, after finding \(\text{pH}\), calculate \(\text{pOH}\) using \(\text{pOH} = 14 - \text{pH}\):(b) \(\text{pOH} = 14 - 2.398 = 11.602\).(d) \(\text{pOH} = 14 - 0.18 = 13.82\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Strong Acids
Chemistry tells us that strong acids are substances that completely dissociate in water to release hydrogen ions, \(\text{H}^+\). This complete dissociation results in a solution with a high concentration of hydrogen ions. The more hydrogen ions there are, the stronger the acidity of the solution.
Some common examples of strong acids include hydrochloric acid (HCl), sulfuric acid (H鈧係O鈧�), and perchloric acid (HClO鈧�). In our exercise, HI (hydroiodic acid) and HClO鈧� (perchloric acid) are both strong acids.
  • Complete Dissociation: Strong acids release \(\text{H}^+\) ions completely upon dissolving.
  • Low \(\text{pH}\): Because they produce a lot of \(\text{H}^+\) ions, their \(\text{pH}\) values are typically very low, under 3.
Understanding strong acids helps us easily calculate their \(\text{pH}\) based on concentration, using the formula: \(\text{pH} = -\log([\text{H}^+])\).
Strong Bases
Strong bases are the chemical counterparts to strong acids. They dissociate entirely in water to produce hydroxide ions, \(\text{OH}^-\). This complete dissociation results in a high concentration of hydroxide ions in the solution, which denotes a high basicity or alkalinity.
Examples of strong bases include sodium hydroxide (NaOH), potassium hydroxide (KOH), and in our case, cesium hydroxide (CsOH) and lithium hydroxide (LiOH).
  • Complete Dissociation: Strong bases release \(\text{OH}^-\) ions completely, increasing basicity.
  • High \(\text{pH}\): Solutions of strong bases typically have high \(\text{pH}\) values, often above 11.
Calculating the \(\text{pOH}\) of strong bases is vital. Start by determining the concentration of \(\text{OH}^-\) ions, then use the formula: \(\text{pOH} = -\log([\text{OH}^-])\).
pH Scale
The \(\text{pH}\) scale is a numerical representation that indicates the acidity or basicity of a solution ranging from 0 to 14.
The scale is logarithmic, which means it changes by tenfold at each step.
  • Acidic Solutions: \(\text{pH} \lt 7\) indicates an acidic environment. The lower the \(\text{pH}\), the stronger the acid.
  • Neutral Solutions: \(\text{pH} = 7\) is neutral, representing pure water.
  • Basic Solutions: \(\text{pH} \gt 7\) denotes a basic or alkaline solution. Higher \(\text{pH}\) values mean stronger bases.
For a comprehensive understanding, \(\text{pH}\) and \(\text{pOH}\) relate, with the equation: \(\text{pH} + \text{pOH} = 14\). This relationship is critical for converting between \(\text{pH}\) and \(\text{pOH}\).
Logarithmic Equations
Logarithmic equations are vital for understanding pH and pOH calculations. In a logarithm, the power to which a base number (commonly 10) is raised gives another number. In chemistry, logarithms provide a simplified way to express exceedingly large or small numbers, like hydrogen and hydroxide ion concentrations.
Logarithmic functions are used to calculate \(\text{pH}\) and \(\text{pOH}\). For instance:
  • \

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Most popular questions from this chapter

A hypothetical weak base has \(K_{b}=5.0 \times 10^{-4}\). Calculate the equilibrium concentrations of the base, its conjugate acid, and \(\mathrm{OH}^{-}\) in a \(0.15 \mathrm{M}\) solution of the base.

Calculate the \(\mathrm{pH}\) and \(\mathrm{pOH}\) of the following solutions. (a) \(0.050 M \mathrm{HCl}\) (b) \(0.024 \mathrm{M} \mathrm{KOH}\) (c) \(0.014 \mathrm{M} \mathrm{HClO}_{4}\) (d) \(1.05 \mathrm{M} \mathrm{NaOH}\)

For each of the following reactions, identify the Br酶nsted-Lowry acids and bases. What are the conjugate acid/base pairs? (a) \(\mathrm{CN}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{HCN}+\mathrm{OH}^{-}\) (b) \(\mathrm{HCO}_{3}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{HS}^{-} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}_{2} \mathrm{~S}\)

Choose from among the labels strongly acidic, weakly acidic, neutral, weakly basic, and strongly basic to estimate the \(\mathrm{pH}\) of the following solutions. (a) \(0.30 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) (b) \(0.25 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) (c) \(0.080 \mathrm{M} \mathrm{HI}\) (d) \(0.12 M \mathrm{LiI}\)

Consider sodium acrylate, \(\mathrm{NaC}_{3} \mathrm{H}_{3} \mathrm{O}_{2} . K_{\mathrm{a}}\) for acrylic acid (its conjugate acid) is \(5.5 \times 10^{-5}\). (a) Write a balanced net ionic equation for the reaction that makes aqueous solutions of sodium acrylate basic. (b) Calculate \(K_{\mathrm{b}}\) for the reaction in (a). (c) Find the \(\mathrm{pH}\) of a solution prepared by dissolving \(1.61 \mathrm{~g} \mathrm{NaC}_{3} \mathrm{H}_{3} \mathrm{O}_{2}\) in enough water to make \(835 \mathrm{~mL}\) of solution.
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