A cylinder at left with balls evenly spaced throughout the cylinder has an arrow leading to a cylinder at right cylinder with ba
2 answers:
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Hello!
To solve this problem we are going to use the Henderson-Hasselbach equation and clear for the molar ratio. Keep in mind that we need the value for Acetic Acid's pKa, which can be found in tables and is 4,76:
![pH=pKa + log ( \frac{[CH_3COONa]}{[CH_3COOH]} )](https://tex.z-dn.net/?f=pH%3DpKa%20%2B%20log%20%28%20%5Cfrac%7B%5BCH_3COONa%5D%7D%7B%5BCH_3COOH%5D%7D%20%29%20)
![\frac{[CH_3COOH]}{[CH_3COONa}= 10^{(pH-pKa)^{-1}}=10^{(4-4,76)^{-1}}=5,75](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BCH_3COOH%5D%7D%7B%5BCH_3COONa%7D%3D%2010%5E%7B%28pH-pKa%29%5E%7B-1%7D%7D%3D10%5E%7B%284-4%2C76%29%5E%7B-1%7D%7D%3D5%2C75%20)
So, the mole ratio of CH₃COOH to CH₃COONa is 5,75
Have a nice day!
To solve this problem we are going to use the Henderson-Hasselbach equation and clear for the molar ratio. Keep in mind that we need the value for Acetic Acid's pKa, which can be found in tables and is 4,76:
So, the mole ratio of CH₃COOH to CH₃COONa is 5,75
Have a nice day!
<u>Answer:</u>
<u>For a:</u> The equilibrium mixture contains primarily reactants.
<u>For b:</u> The equilibrium mixture contains primarily products.
<u>Explanation:</u>
There are 3 conditions:
- When
; the reaction is product favored.
- When
; the reaction is reactant favored.
- When
; the reaction is in equilibrium.
For the given chemical reactions:
- <u>For a:</u>
The chemical equation follows:
The expression of for above reaction follows:
As, , the reaction will be favored on the reactant side.
Hence, the equilibrium mixture contains primarily reactants.
- <u>For b:</u>
The chemical equation follows:
The expression of for above reaction follows:
As, , the reaction will be favored on the product side.
Hence, the equilibrium mixture contains primarily products.