The change in pH of a 1.00 L of a buffered solution preparing by mixing 0.50 M acetic acid (Ka = 1.8 x 10^-5) and 0.50 M sodium acetate when 0.010 mole of NaOH is added is 4.75
when the same amount 0.010 mole of NaOH was added to 1.00 L of water the pH = 12
Explanation:
given that:
concentration of acetic acid = 0.50 M
Concentration of base sodium acetate = 0.50 M
ka = 1.8 x 10^-5)
pka = -log [ka]
pka = 4.74
From Henderson-Hasselbalch Equation:
pH = pKa + log ![\frac{[base]}{[acid]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D)
pH = 4.74 + Log ![\frac{[0.5]}{[0.5]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B0.5%5D%7D%7B%5B0.5%5D%7D)
pH = 4.74 + 0
pH = 4.74
Number of moles of NaOH = 0.010 moles
volume 1 litre
molarity = 0.010 M
Moles of acetic acid and sodium acetate before addition of NaOH
FORMULA USED:
molarity = ![\frac{number of moles}{volume in litres}](https://tex.z-dn.net/?f=%5Cfrac%7Bnumber%20of%20moles%7D%7Bvolume%20in%20litres%7D)
acetic acid,
0.5 = number of moles
0.5 is the number of moles of sodium acetate.
number of moles of NaOH 0.010 moles
NaOH reacts in 1:1 molar ratio with acetic acid so
number of moles in acetic acid = 0.5 - 0.010 = 0.49
number of moles in sodium acetate = 0.5 +0.010 = 0.51
new pH
pH = pKa + log ![\frac{[base]}{[acid]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D)
pH= 4.74 + log[0.51] - log[0.49]
pH= 4.75
PH of NaOH of 0.01 M (BASE)
pOH = -Log[0.01]
pOH = 2
pH can be calculated as
14= pH +pOH
pH= 14-2
pH = 12
A visual representation of covalent bonding which represents the valence shell electrons in the molecule is said to be a Lewis structure. The lines represents the shared electron pairs and dots represents the electrons that are not involved in the bonding i.e lone pairs.
Number of valence electrons in each atom:
For Carbon,
= 4
For Hydrogen,
= 1
For Nitrogen,
= 5
The Lewis structure of
is shown in the attached image.
The formula of calculating formula charge =
-(1)
where, F.C is formal charge, V.E is number of valence electrons, N.E is number of non-bonding electrons and B.E is number of bonding electrons.
Now, calculating the formal charge:
For
on left side:
![FC = 4 - 0- \frac{8}{2} = 0](https://tex.z-dn.net/?f=FC%20%3D%204%20-%200-%20%5Cfrac%7B8%7D%7B2%7D%20%3D%200)
For
:
![FC = 5 - 0- \frac{8}{2} = +1](https://tex.z-dn.net/?f=FC%20%3D%205%20-%200-%20%5Cfrac%7B8%7D%7B2%7D%20%3D%20%2B1)
For
on right side:
![FC = 4 - 2- \frac{6}{2} = -1](https://tex.z-dn.net/?f=FC%20%3D%204%20-%202-%20%5Cfrac%7B6%7D%7B2%7D%20%3D%20-1)
The formula charge of each atom other than hydrogen is shown in the attached image.
Answer:
The answer is 17.03052. We assume you are converting between grams Ammonia and mole.
A complex, ML₆²⁺, is violet. The same metal forms a complex with another ligand, Q, that creates a weaker field. MQ₆²⁺, be expected to show green color.
<h3>What is spectrochemical series?</h3>
The ligands (attachment to a metal ion) are listed in the spectrochemical series according to the strength of their field. The series has been created by superimposing several sequences discovered through spectroscopic research because it is impossible to produce the full series by examining complexes with a single metal ion. The halides are referred to be weak-field ligands whereas the ligands cyanide and CO are strong-field ligands. Medium field effects are claimed to be produced by ligands like water and ammonia.
To know more about the spectrochemical series, visit:
brainly.com/question/27892620
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