Conjugate base of Propanoic acid (
is propanoate where -COOH group gets converted to -CO
. The structure of conjugate base of Propanoic acid is shown in the diagram.
The 
above which 90% of the compound will be in this conjugate base form can be determined using Henderson's equation as propanoic acid is weak acid and it can form buffer solution on reaction with strong base.
= 
+ log![\frac{[Conjugate base]}{[acid]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BConjugate%20base%5D%7D%7B%5Bacid%5D%7D)
=4.9+log
=5.85
As 90% conjugate base is present, so propanoic acid present 10%.
LiOH is soluble. Na2CO3 is soluble. Cu(OH)2 is insoluble.
0.000132 g of hydrated sodium borate (Na₂B₄O₇ · 10 H₂O)
Explanation:
First we need to find the number of moles of sodium borate (Na₂B₄O₇) in the solution:
molar concentration = number of moles / volume (L)
number of moles = molar concentration × volume (L)
number of moles of Na₂B₄O₇ = 0.1 × 0.5 = 0.05 moles
We know now that we need 0.05 moles of hydrated sodium borate (Na₂B₄O₇ · 10 H₂O) to make the solution.
Now to find the mass of hydrated sodium borate we use the following formula:
number of moles = mass / molar weight
mass = number of moles × molar weight
mass of hydrated sodium borate = 0.05 / 381 = 0.000132 g
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molar concentration
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By circle fraction additional number for the number
