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olga55 [171]
3 years ago
11

25 POINTS!!!

Chemistry
1 answer:
Verdich [7]3 years ago
7 0

Answer:

1`0

Explanation:

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Please Help! Thank You
ra1l [238]

Answer:

oooooooooooooooooh i dont know this one

Explanation:

im confuzzled

6 0
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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I THINK this is the answer: 4809.35999999786. (found it on a website)

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