Calculate [OH-] for a solution formed by adding 4.80 mL of 0.180 M KOH to 17.0 mL of 6.5*10-2 M Ca (OH) 2.

[OH⁻] = 0.167 mol/L

Explanation:

Let's apply Kps to solve this, as the Ca (OH) ₂is only slighty solube in water. We imagine that Temperature of working, is 25°C.

Ca (OH) ₂ → Ca²⁺ + 2OH⁻ Kps = 5.02x10⁻⁶ (25°C)

s 2s

Kps = 2s³

5.02x10⁻⁶ = 2s³

5.02x10⁻⁶ / 2 = s³ → ∛ (5.02x10⁻⁶ / 2) = s

s = 0.013

So [OH⁻] = 0.027 mol/L

Now, we add KOH

[OH⁻] = 3.74x10⁻³ moles / 0.0218 L = 0.14 mol/L

Finally, the concentration of [OH⁻] will be the sum of, hydroxide from Ca (OH) ₂ and KOH

0.027 mol/L + 0.14 mol/L = 0.167 mol/L