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Ray Of Light [21]
3 years ago
11

Write the Ka expression for an aqueous solution of hydrofluoric acid: (Note that either the numerator or denominator may contain

more than one chemical species. Enter the complete numerator in the top box and the complete denominator in the bottom box. Remember to write the hydronium ion out as , and not as ) Ka
Chemistry
1 answer:
damaskus [11]3 years ago
4 0

Answer:

Ka = ( [H₃O⁺] . [F⁻] ) / [HF]

Explanation:

HF is a weak acid which in water, keeps this equilibrium

HF (aq)  +  H₂O (l)  ⇄  H₃O⁺ (aq)  +  F⁻ (aq)      Ka

2H₂O (l)  ⇄  H₃O⁺ (l)  +  OH⁻ (aq)   Kw

HF is the weak acid

F⁻ is the conjugate stron base

Let's make the expression for K

K = ( [H₃O⁺] . [F⁻] ) / [HF] . [H₂O]

K . [H₂O] = ( [H₃O⁺] . [F⁻] ) / [HF]

K . [H₂O] = Ka

Ka, the acid dissociation constant, includes Kwater.

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What is the molarity of a solution of 58.7 grams of MgCl2 in 359 ml of solution?
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Answer:

1.72 M

Explanation:

Molarity is the molar concentration of a solution. It can be calculated using the formula a follows:

Molarity = number of moles (n? ÷ volume (V)

According to the information provided in this question, the solution has 58.7 grams of MgCl2 in 359 ml of solution.

Using mole = mass/molar mass

Molar mass of MgCl2 = 24 + 35.5(2)

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