Question

Calculate pH of 0.10 M hydrochloric acid (why no Ka?)

Calculate the pH of a 0.100 M calcium hydroxide (hint: write the formula first & rxn in water)

Calculate the pH and % ionization of a 0.05 M hydrofluoric acid (ka = 8.0 x 10 ^-4) .

What is the hydrogen ion concentration for a hydrochloric acid solution that has a pH of 2.30?

Answer 1

HCl is a strong acid. That is complete ionization takes place in a solution. Thus we are not taking Ka.

0.1 m HCl yield 0.1 M protons

pH = -log [H+] = -log [0.1] = 1

2.

Ca(OH)2 give Ca2+ + 2OH-

one Ca(OH)2 yields 2 OH-

0.1M yields 0.2 M OH-

pOH = -log [OH-] = -log [0.2] = 0.6989

pH = 14-0.6989= 13.3

3.

HF in equilibrium with H+ and F-

at equilibrium, if X is the dissociation constant and C os the concentration left.

C(1-X) will be the remaining HF

CX will be concentrations of H+ and OH-

equilibrium constant Ka = CX x cX/C(1-X)

since xis very smaller 1-x can be approximated to 1

then

X = sqrt (ka/c) = (8 x10^-4/0.05) = 0.126

'Concentration of proton = CX = 0.05 x 0.126 = 0.0063 M

pH = -log [H+] = 2.2

4.

pH = -log [H+]

[H+] = antilog [-pH] = 0.005 M