Answer:
1) The ranked solutions from the lowest pH to the highest pH are:
HCl < HF < HClO < NaCl
2) The ranked solutions from the lowest pH to the highest pH are:
HCl < CH₃COOH < NaCl < NH₃ < KOH
Explanation:
1) HCl is a strong acid because the concentration of the undissociated species HCl is too low to be measured. A 0.1 M solution of HCl is pH=1
pH=-log[H₃O⁺]=-log[0.1]=1
HF and HClO are weak acids because they partially dissociates when they are dissolved in a solvent.
HF pKa =3.17 ⇒ Ka=[tex]10^{-3.17}[/tex]=6.761×10⁻⁴
pH= [tex]-log[\sqrt{c*Ka} ][/tex]= -log[√(0.1×6.761×10⁻⁴)]=2.1
HClO pKa=7.49 ⇒ Ka=[tex]10^{-7.49}[/tex]=3.236×10⁻⁸
pH= [tex]-log[\sqrt{c*Ka} ][/tex]= -log[√(0.1×3.236×10⁻⁸)]=4.2
NaCl is a neutral salt. The pH of the solution is 7.
[tex]pH_{HCl}=1 < pH_{HF}=2.1 < pH_{HClO}=4.2 < pH_{NaCl}=7[/tex]
HCl < HF < HClO < NaCl
2) HCl is a strong acid because the concentration of the undissociated species HCl is too low to be measured. A 0.1 M solution of HCl is pH=1
KOH is a strong base because the concentration of the undissociated species KOH is too low to be measured. A 0.1 M solution of KOH is pOH=1 ⇒ pH= 14 - pOH = 13
CH₃COOH (acetic acid) is a weak acid whose pKa=4.76
Ka=[tex]10^{-4.76}[/tex]=1.738×10⁻⁵
pH= [tex]-log[\sqrt{c*Ka} ][/tex]= -log[√(0.1×1.738×10⁻⁵)]=2.88
NH₃ is a weak base whose pKb=5 ⇒ ⇔Kb=1×10⁻⁵
NH₃ + H₂O ⇔ NH₄⁺ + OH⁻
[OH⁻]=[tex]\sqrt{c*Kb}[/tex]= 1×10⁻³
pOH=-log[1×10⁻³]=3
pH=14-pOH= 11
Hence, the rank should be:
[tex]pH_{HCl}=1 < pH_{CH_{3}COOH}=2.88 < pH_{NaCl}=7 < pH_{NH_3}=11 < pH_{KOH}=13 [/tex]
HCl < CH₃COOH < NaCl < NH₃ < KOH
