Question

Calculate the pH (to two decimal places) for the following:

6.89×10^-5 M Sr(OH)₂

Calculate the pOH (to two decimal places) for the following:
a) 3.36×10^-5 M HCl

b) 2.69×10^-2 M KH

Answer 1

6.89×10-5 M Sr(OH)2

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

[OH-] = 6.89*10^-5)*2 = 0.0001378

pOH = -log(OH) = -log(0.0001378) = 3.86

pH = 14-pOH = 14-3.86 = 10.14

2)

Calcualte pOH

pH of HCL:

pH = -loG(H+) = -log(HCl) = -log(3.36*10^-5) = 4.473

pOH = 14-pH = 14-4.47 = 9.53

b)

NOTE: you are probably searching for KOH, which is abase

KH is potassium hydride and is not typically used in examples of acid/base

Then

pOH = -log(OH-)

KOH = K+ and OH-

[OH-] = 2.69*10^-2

pOH = -log(2.69*10^-2) = 1.57024