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Ray Of Light [21]
2 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]2 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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In a sulphuric acid (h2so4) - sodium hydroxide (naoh) acid-base titration, 17.3 ml of 0.126 m naoh is needed to neutralize 25 ml
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The balanced equation for the neutralisation reaction is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
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therefore if 25 mL contains - 0.00109 mol 
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3 years ago
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2 years ago
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3 years ago
If acetic acid is the only acid that vinegar contains (ka=1.8×10−5), calculate the concentration of acetic acid in the vinegar.
kicyunya [14]
Ethanoic (Acetic) acid is a weak acid and do not dissociate fully. Therefore its equilibrium state has to be considered here.

CH_{3}COOH \ \textless \ ---\ \textgreater \   H^{+} + CH_{3}COO^{-}

In this case pH value of the solution is necessary to calculate the concentration but it's not given here so pH = 2.88 (looked it up)

pH = 2.88 ==> [H^{+}]  = 10^{-2.88} =  0.001 moldm^{-3}

The change in Concentration Δ [CH_{3}COOH]= 0.001 moldm^{-3}


                                  CH3COOH          H+           CH3COOH    
Initial  moldm^{-3}                      x           0                     0
                                                                                                                       
Change moldm^{-3}        -0.001            +0.001           +0.001
                                                                                                       
Equilibrium moldm^{-3}      x- 0.001      0.001             0.001
                                                                              

Since the k_{a} value is so small, the assumption 
[CH_{3}COOH]_{initial} = [CH_{3}COOH]_{equilibrium} can be made.

k_{a} = [tex]= 1.8*10^{-5}  =  \frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} =  \frac{0.001^{2}}{x}

Solve for x to get the required concentration.

note: 1.)Since you need the answer in 2SF don&t round up values in the middle of the calculation like I've done here.

         2.) The ICE (Initial, Change, Equilibrium) table may come in handy if you are new to problems of this kind

Hope this helps! 



8 0
3 years ago
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