Question

How would i go about answering this?

Select the range the PH will fall into for each .1M solution.

a. MnCl3 <1 1 between 1&7 7 between 7&13 13 >13

b. CH3CO2H <1 1 between 1&7 7 between 7&13 13 >13

c. KOH <1 1 between 1&7 7 between 7&13 13 >13

d. NaF <1 1 between 1&7 7 between 7&13 13 >13

Answer 1

MnCl3 reacts with water as MnCl3 + H2O --> MnCl2 + MnO2 + HCl.

The reaction of MnCl3 with water produced HCl, which is strong acid. Hence the pH will be less than <1

HCl ionizes completely in water. Hence [H+]=0.1M, pH=1,

2. Acetic acid is weak acid Ka= 1.8*10^-5, the ionization of acetic acid (HA) is

HA+H2O ------->H3O+A-, Ka= [H3O+] [A-]/ [HA]

let x= drop in concentration of 0.1M HA

at equilibrium [HA]=0.1-x, [A-]= [H3O+]=x, Ka= x²/(0.1-x)= 1.8*10^-5, when solved using excel, x= 1.35*10-3

pH= -log [H3O+]= 2.87 lies between 1and 7.

3. KOH is strong base . Hence it completely dissociates in water. [OH-]= 0.1, pOH= -log (0.1)= 1, pH= 14-pOH= 13

pH will be 13.

4. NaF is base formed from reaction of weak acid HF and strong base NaOH.

NaF+ H2O ------->HF+OH-

Kb= [HF] [OH-]/[NaF], Ka of HF= 6.6*10-4, Kb= 10^-14/ (6.6*10^-4)= 1.52*10^-11

let x= drop in concentration of 0.1M NaF to reach equilibrium

at Equilibrium [NaF]=0.1-x, [HF]= [OH-]=x

hence x²/(0.1-x)= 1.52*10^-11, when solved using excel, x=1.23*10^-6, pOH= -log (1.23*10^-6)= 5.91, pH= 14-5.91= 8.09

between 7 and 13.

2.