Question

A solution contains 0.1252 grams of strong base calcium hydroxide completely dissolved in 1255 ml of pure water. What is the pH of this solution?

Answer 1

Molar mass of Ca(OH)2 = 1*MM(Ca) + 2*MM(O) + 2*MM(H)

= 1*40.08 + 2*16.0 + 2*1.008

= 74.096 g/mol

mass of Ca(OH)2 = 0.1252 g

we have below equation to be used:

number of mol of Ca(OH)2,

n = mass of Ca(OH)2/molar mass of Ca(OH)2

=(0.1252 g)/(74.096 g/mol)

= 1.69*10^-3 mol

volume , V = 1255 mL

= 1.255 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 1.69*10^-3/1.255

= 1.346*10^-3 M

So,

[Ca(OH)2] = 1.346*10^-3 M

[OH-] = 2*[Ca(OH)2]

= 2*1.346*10^-3 M

= 2.692*10^-3 M

we have below equation to be used:

pOH = -log [OH-]

= -log (2.692*10^-3)

= 2.57

we have below equation to be used:

PH = 14 - pOH

= 14 - 2.57

= 11.43

Answer: 11.43