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Chapter 14: Problem 158

Rank the following \(0.10 M\) solutions in order of increasing \(\mathrm{pH}\). a. HI, HF, NaF, NaI b. \(\mathrm{NH}_{4} \mathrm{Br}, \mathrm{HBr}, \mathrm{KBr}, \mathrm{NH}_{3}\) c. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{NO}_{3}, \mathrm{NaNO}_{3}, \mathrm{NaOH}, \mathrm{HOC}_{6} \mathrm{H}_{5}, \mathrm{KOC}_{6} \mathrm{H}_{5}\), \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{HNO}_{3}\)

Short Answer

Expert verified
Increasing pH for each set of solutions: Set a: HI < HF < NaI < NaF Set b: HBr < NH4Br < KBr < NH3 Set c: HNO3 < C6H5NH3NO3 < HOC6H5 < NaNO3 < C6H5NH2 < KOC6H5 < NaOH

Step by step solution

01

Set a: Identify the nature of the substances

HI: Strong acid (hydrogen iodide) HF: Weak acid (hydrofluoric acid) NaF: Salt derived from weak acid (HF) and strong base (NaOH), produces a basic solution NaI: Salt derived from strong acid (HI) and strong base (NaOH), produces a neutral solution
02

Set a: Rank the solutions

Increasing pH: HI < HF < NaI < NaF
03

Set b: Identify the nature of the substances

NH4Br: Salt derived from weak base (NH3) and strong acid (HBr), produces an acidic solution HBr: Strong acid (hydrobromic acid) KBr: Salt derived from strong acid (HBr) and strong base (KOH), produces a neutral solution NH3: Weak base (ammonia)
04

Set b: Rank the solutions

Increasing pH: HBr < NH4Br < KBr < NH3
05

Set c: Identify the nature of the substances

C6H5NH3NO3: Salt derived from weak base (C6H5NH2) and strong acid (HNO3), produces an acidic solution NaNO3: Salt derived from strong acid (HNO3) and strong base (NaOH), produces a neutral solution NaOH: Strong base (sodium hydroxide) HOC6H5: Weak acid (phenol) KOC6H5: Salt derived from weak acid (HOC6H5) and strong base (KOH), produces a basic solution C6H5NH2: Weak base (aniline) HNO3: Strong acid (nitric acid)
06

Set c: Rank the solutions

Increasing pH: HNO3 < C6H5NH3NO3 < HOC6H5 < NaNO3 < C6H5NH2 < KOC6H5 < NaOH

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Most popular questions from this chapter

Acrylic acid \(\left(\mathrm{CH}_{2}=\mathrm{CHCO}_{2} \mathrm{H}\right)\) is a precursor for many important plastics. \(K_{\mathrm{a}}\) for acrylic acid is \(5.6 \times 10^{-5}\) a. Calculate the \(\mathrm{pH}\) of a \(0.10-M\) solution of acrylic acid. b. Calculate the percent dissociation of a \(0.10-M\) solution of acrylic acid. c. Calculate the \(\mathrm{pH}\) of a \(0.050-M\) solution of sodium acrylate \(\left(\mathrm{NaC}_{3} \mathrm{H}_{3} \mathrm{O}_{2}\right)\)

Consider \(0.10 M\) solutions of the following compounds: \(\mathrm{AlCl}_{3}\) \(\mathrm{NaCN}, \mathrm{KOH}, \mathrm{CsClO}_{4}\), and NaF. Place these solutions in order of increasing \(\mathrm{pH}\).

Use the Lewis acid-base model to explain the following reaction. $$ \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{CO}_{3}(a q) $$

Consider \(1000 . \mathrm{mL}\) of a \(1.00 \times 10^{-4}-M\) solution of a certain acid HA that has a \(K_{\mathrm{a}}\) value equal to \(1.00 \times 10^{-4}\). How much water was added or removed (by evaporation) so that a solution remains in which \(25.0 \%\) of HA is dissociated at equilibrium? Assume that HA is nonvolatile.

For solutions of the same concentration, as acid strength increases, indicate what happens to each of the following (increases, decreases, or doesn't change). a. \(\left[\mathrm{H}^{+}\right]\) d. \(\mathrm{pOH}\) b. \(\mathrm{pH}\) e. \(K_{\mathrm{a}}\) c. \(\left[\mathrm{OH}^{-}\right]\)
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