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melamori03 [73]
2 years ago
8

Net ionic equation for Calcium hydroxide reacting with ammonium Chloride?

Chemistry
2 answers:
Anton [14]2 years ago
8 0

Explanation:

ca (OH)2 + 2NH4CL------> CACL2 + 2NH3 + 2H2O

VMariaS [17]2 years ago
3 0

Answer:

Ca(OH)₂ (aq) + 2NH₄Cl (aq) → CaCl₂ (aq) + 2NH₃ (g) + 2H₂O (l)

Ca²⁺ + 2OH⁻ + 2NH₄⁺ + 2Cl⁻ → Ca⁺² + 2Cl⁻ + 2NH₃ (g) + 2H₂O (l)

2OH⁻ + 2NH₄⁺ → 2NH₃ (g) + 2H₂O (l)

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If the K a Ka of a monoprotic weak acid is 7.3 × 10 − 6 , 7.3×10−6, what is the pH pH of a 0.40 M 0.40 M solution of this acid?
olga_2 [115]

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The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.

In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:

                          HA                                   H₃O⁺                          A⁻          

Initial, M             0.40                                   0                              0

Change , M          -x                                     +x                            +x

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7.3 x 10⁻⁶ = x² / 0.40  ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³

[H₃O⁺] = 1.71 x 10⁻³

Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.

pH = - log ( 1.71 x 10⁻³ ) = 3.8

Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.

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Hope I helped!

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