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madreJ [45]
3 years ago
12

If kb for nx3 is 4.0×10−6, what is the poh of a 0.175 m aqueous solution of nx3?

Chemistry
1 answer:
Elena-2011 [213]3 years ago
7 0
<h3><u>Answer;</u></h3>

pOH = 3.08

<h3><u>Explanation;</u></h3>

NX3 + H2O <----> NHX3+ + OH-  

Kb = 4.0 x 10^-6

Kb = c(NH₄⁺) · c(OH⁻) ÷ c(NH₃).

c(NH₄⁺) = c(OH⁻) = x.

x² = Kb · c(NH₃)

x² = 4.0 × 10⁻⁶ × 0.175 = 7.0 × 10⁻⁷.

x = c(OH⁻) = √(7.0 × 10⁻⁷)

    = 8.367 × 10⁻⁴

pOH = -log(c(OH⁻))

        =- log ( 8.367 × 10⁻⁴)

        <u>= 3.08</u>

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Step 4) Estimate and submit your answer

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If the pka of hcho2 is 3.74 and the ph of an hcho2/nacho2 solution is 3.89, it means that
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Hi, the question has some missing part. The complete question should be " If the pK_{a} of HCHO_{2} is 3.74 and the pH of an HCHO_{2}/NaCHO_{2} solution is 3.89, it means that -

(a) [HCHO_{2}]=[NaCHO_{2}]

(b) [HCHO_{2}]> [NaCHO_{2}]

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Answer: Option (c) is true.

Explanation:

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Here, pH = 3.89, pK_{a} of HCHO_{2} = 3.74

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So,  [CHO_{2}^{-}]> [HCHO_{2}]

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