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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.

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Answer:

[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

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