Additional Practice Questions 6. If you dissolve 2.31 g of barium hydroxide into 250 mL of water and the barium hydroxide is fully ionized, what will
your resulting pH be?

pH of solution = 13.033

Given

2.31 g Ba(OH)₂

250 ml water

Required

pH of solution

Solution

Barium hydroxide is fully ionized, means that Ba(OH)₂ is a strong base

So we use a strong base formula to find the pH

[OH ⁻] = b. Mb where

b = number of OH⁻ /base valence

Mb = strong base concentration

Molarity of Ba(OH)₂(MW=171.34 g/mol) :

Ba(OH)₂ ⇒ Ba²⁺ + 2OH⁻(b=valence=2)

[OH⁻]= 2 . 0.054

[OH⁻] = 0.108

pOH= - log 0.108

pOH=0.967

pOH+pH=14

pH=14-0.967

pH=13.033

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2