100ml of 0.05M of NaOH is added to 0.5g of Ca(OH)2. find the [OH] (from Ca(OH)2)

To find the concentration of hydroxide ions ([OH-]) from calcium hydroxide (Ca(OH)2), we can use the stoichiometry of the balanced chemical equation and the given information.

First, let's write the balanced chemical equation for the reaction between sodium hydroxide (NaOH) and calcium hydroxide (Ca(OH)2:

Ca(OH)2 + 2NaOH → Ca(OH)2 + 2NaOH

From the equation, we can see that 1 mole of Ca(OH)2 reacts with 2 moles of NaOH. This means that the number of moles of NaOH required to react with 1 mole of Ca(OH)2 is 2.

Given that the volume of NaOH solution is 100 mL (or 0.1 L) and the concentration of NaOH is 0.05 M, we can calculate the number of moles of NaOH used:

Moles of NaOH = Volume (L) x Concentration (M)
= 0.1 L x 0.05 M
= 0.005 moles

Now, let's convert the mass of Ca(OH)2 given (0.5 g) to moles. The molar mass of Ca(OH)2 is:

Ca = 40.08 g/mol
O = 16.00 g/mol (there are two oxygen atoms)
H = 1.01 g/mol (there are two hydrogen atoms)

Molar mass of Ca(OH)2 = (40.08 g/mol) + 2(16.00 g/mol) + 2(1.01 g/mol)
= 40.08 g/mol + 32.00 g/mol + 2.02 g/mol
= 74.10 g/mol

Moles of Ca(OH)2 = Mass (g) / Molar mass (g/mol)
= 0.5 g / 74.10 g/mol
≈ 0.00674 moles

Since the reaction is 1:1 between Ca(OH)2 and OH-, we can conclude that 0.00674 moles of OH- is formed.

Now, let's find the total volume in liters by adding the volume of NaOH solution (0.1 L) to the volume of water formed by the dissolution of Ca(OH)2. For Ca(OH)2, the volume of water is 2 moles of OH- ions:

Volume of water = 2 moles x (1 L / 55.55 moles) (The molar volume of water is approximately 55.55 L/mol)
≈ 0.036 L

Total volume = Volume of NaOH solution + Volume of water
= 0.1 L + 0.036 L
= 0.136 L

Finally, we can calculate the concentration of OH- ions ([OH-]):

[OH-] = Moles of OH- ions / Total volume (L)
= 0.00674 moles / 0.136 L
≈ 0.0496 M

Therefore, the concentration of hydroxide ions ([OH-]) from Ca(OH)2 is approximately 0.0496 M.