the consentration of Sr(OH)2 I
S 0.02 WHAT IS THE CONSENTRATION OF OH ASSUMING COMPLETE DISSOCIATION
Concentration is the word you need.
if the molar concentration of Sr(OH)2 is .02M, then you have with complete dissociation .02M of Sr++, and .04M of OH-. Now why did the OH double? The reason is that in each molecule of Sr(OH)2 you had two OH ions.
To determine the concentration of OH- assuming complete dissociation of Sr(OH)2, we need to consider that Sr(OH)2 dissociates into two OH- ions for each formula unit.
Given the concentration of Sr(OH)2 as 0.02 M, we can convert this to the concentration of OH- using stoichiometry.
First, we determine the moles of Sr(OH)2 by multiplying the concentration (0.02 M) by the volume of the solution used.
Next, we use the stoichiometry of the reaction:
1 mole of Sr(OH)2 produces 2 moles of OH- ions.
Therefore, the number of moles of OH- ions is twice the moles of Sr(OH)2.
Finally, to find the concentration of OH-, divide the moles of OH- ions by the volume of the solution used.
Let's assume we used a volume of 100 mL for the solution. The steps are as follows:
Step 1: Calculate moles of Sr(OH)2:
Moles of Sr(OH)2 = concentration of Sr(OH)2 × volume of solution (in liters)
= 0.02 M × 0.100 L
= 0.002 moles
Step 2: Calculate moles of OH- ions:
Moles of OH- ions = 2 × moles of Sr(OH)2
= 2 × 0.002 moles
= 0.004 moles
Step 3: Calculate concentration of OH- ions:
Concentration of OH- = moles of OH- ions / volume of solution (in liters)
= 0.004 moles / 0.100 L
= 0.04 M
Therefore, assuming complete dissociation of Sr(OH)2, the concentration of OH- is 0.04 M.