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borishaifa [10]
3 years ago
14

What is the pH of a 9.50×10−2 M ammonia solution??

Chemistry
1 answer:
Darina [25.2K]3 years ago
4 0
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻

c=9,50*10⁻² mol/L

Ammonia solution is a weak electrolyte.
The dissociation constant of ammonium hydroxide is:
K=1.78*10⁻⁵

[OH⁻]=√(Kc)

pH=14-pOH=14+lg[OH]⁻
pH=14+lg(√(Kc))

pH=14+lg(√(1.78*10⁻⁵*9.50*10⁻²))=8.23
pH=8.23
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Colligative properties calculations are used for this type of problem. Calculations are as follows:<span>
</span>

<span>ΔT(freezing point)  = (Kf)m
ΔT(freezing point)  = 1.86 °C kg / mol (0.705)
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<span>Hope this answers the question. Have a nice day.</span>

6 0
3 years ago
Find the number of moles of barium iodide if you have 5.23 x 1024 formula units of Bal2
Oksi-84 [34.3K]

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8.68 moles of BaI₂

Explanation:

Given data:

Number of moles of BaI₂ = ?

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By using Avogadro number,

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5.23× 10²⁴ formula units of BaI₂ × 1 mol / 6.022× 10²³ formula units

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7 0
2 years ago
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8 0
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kumpel [21]
3501.75

is the answer

i think

(your welcome)
7 0
3 years ago
Read 2 more answers
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