Question

a) Caluclate the PH and molaraties of all the species in a 0.150M ammonia solution. Ka=5.55x10^-10

b) Calculate the original molarity of strontium hydroxide if the PH is 12.30.

Answer 1

a)

[NH3] = 0.15

NH3 + H2O <--> NH4+ + OH-

Ka = [NH4+][OH-] / [NH3]

5.55*10^-10 = (x*x)/(0.15-x)

solve for x

x = 2.35*10^-5

[OH-] = x = 2.35*10^-5

pOH = -log(OH) = -log(2.35*10^-5) =4.6289

pH = 14- pOH = 14- 4.6289= 9.3711

[OH-] = [NH4+] = 2.35*10^-5

[NH3] = 0.15 - 2.35*10^-5= 0.1499765

b)

Sr(OH)2 --> Sr+2 and 2OH-

then

pOH = -log(OH)

pOH = 14-pH = 14-12.3 = 1.7

[OH-] = 10^-1.7 = 0.01995

[OH-] = 0.01995

Ratio is 2:1 then

0.01995/2 = 0.009975 Sr(OH)2