Rank the following 0.10 m solutions in order of increasing ph (lowest ph to highest ph): hcl hf nacl hoclrank the following 0.10 m solutions in order of increasing ph: hc2h3o2 nacl nh3 hcl koh

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==6.761×10⁻⁴

pH= = -log[√(0.1×6.761×10⁻⁴)]=2.1

HClO pKa=7.49 ⇒ Ka==3.236×10⁻⁸

pH= = -log[√(0.1×3.236×10⁻⁸)]=4.2

NaCl is a neutral salt. The pH of the solution is 7.

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==1.738×10⁻⁵

pH= = -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⁻]== 1×10⁻³

pOH=-log[1×10⁻³]=3

pH=14-pOH= 11

Hence, the rank should be:

HCl < CH₃COOH < NaCl < NH₃ < KOH

50.23 i believe is correct but don't quote me on that.