the equation demonstrates a reaction of sodium chloride and fluorine. 2NaCl+F2=2NaF+Cl2. in order for the reaction to proceed, f
1 answer:
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Ethanoic (Acetic) acid is a weak acid and do not dissociate fully. Therefore its equilibrium state has to be considered here.
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^{+}]](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D)
= 
= 0.001 
The change in Concentration Δ![[CH_{3}COOH]](https://tex.z-dn.net/?f=%5BCH_%7B3%7DCOOH%5D)
= 0.001
CH3COOH H+ CH3COOH
Initial 
0 0
Change -0.001 +0.001 +0.001
Equilibrium - 0.001 0.001 0.001
Since the
value is so small, the assumption
![[CH_{3}COOH]_{initial} = [CH_{3}COOH]_{equilibrium}](https://tex.z-dn.net/?f=%5BCH_%7B3%7DCOOH%5D_%7Binitial%7D%20%3D%20%5BCH_%7B3%7DCOOH%5D_%7Bequilibrium%7D)
can be made.
![k_{a} = [tex]= 1.8*10^{-5} = \frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} = \frac{0.001^{2}}{x}](https://tex.z-dn.net/?f=%20k_%7Ba%7D%20%3D%20%5Btex%5D%3D%201.8%2A10%5E%7B-5%7D%20%20%3D%20%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5BCH_%7B3%7DCOO%5E%7B-%7D%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D%20%3D%20%20%5Cfrac%7B0.001%5E%7B2%7D%7D%7Bx%7D%20)
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!
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 ==>
The change in Concentration Δ
CH3COOH H+ CH3COOH
Initial
Change
Equilibrium
Since the
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!