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Svet_ta [14]
3 years ago
10

Calculate the ph of a 0.005 m solution of potassium oxide k2o

Chemistry
1 answer:
Alecsey [184]3 years ago
3 0
First, we have to see how K2O behaves when it is dissolved in water:

K2O + H20 = 2 KOH

According to reaction K2O has base properties, so it forms a hydroxide in water.
For the reaction next relation follows:

c(KOH) : c(K2O) = 1 : 2

So,

c(KOH)= 2 x c(K2O)= 2 x 0.005 = 0.01 M = c(OH⁻)

Now we can calculate pH:

pOH= -log c(OH⁻) = -log 0.01 = 2

pH= 14-2 = 12




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Suppose 6.63g of zinc bromide is dissolved in 100.mL of a 0.60 M aqueous solution of potassium carbonate. Calculate the final mo
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Answer:

[Zn²⁺] = 4.78x10⁻¹⁰M

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ZnBr₂(aq) + K₂CO₃(aq) → ZnCO₃(s) + 2KBr(aq)

The zinc added produce the insoluble ZnCO₃ with Ksp = 1.46x10⁻¹⁰:

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<em>Moles ZnB₂ (Molar mass: 225.2g/mol) = Moles Zn²⁺:</em>

6.63g ZnBr₂ * (1mol / 225.2g) = 0.02944moles Zn²⁺

<em>Moles K₂CO₃ = Moles CO₃²⁻:</em>

0.100L * (0.60mol/L) = 0.060 moles CO₃²⁻

Moles CO₃²⁻ in excess: 0.0600moles CO₃²⁻ - 0.02944moles =

0.03056moles CO₃²⁻ / 0.100L = 0.3056M = [CO₃²⁻]

Replacing in Ksp expression:

1.46x10⁻¹⁰ = [Zn²⁺] [0.3056M]

<h3>[Zn²⁺] = 4.78x10⁻¹⁰M</h3>

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3 years ago
How many liters of water would be needed to make a 0.50 M solution from 100.0 g of NaOH?
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