5f2 (g) + 2nh3 (g) → n2f4 (g) + 6hf (g) if you have 58.5g nh3, how many grams of f2 are required for a complete reaction? how ma
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Answer: The value of is 0.0012 M and
is
.
Explanation:
pH is the negative logarithm of concentration of hydrogen ion.
It is given that pH is 2.89. So, the value of concentration of hydrogen ions is calculated as follows.
The relation between pH and pOH value is as follows.
pH + pOH = 14
0.0012 + pOH = 14
pOH = 14 - 0.0012 = 13.99
Now, pOH is the negative logarithm of concentration of hydroxide ions.
Hence, is calculated as follows.
Thus, we can conclude that the value of is 0.0012 M and
is
.
PH = pKa + log ![\frac{[base]}{[Acid]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5Bbase%5D%7D%7B%5BAcid%5D%7D%20)
Acid is HC₂H₃O₂ and conjugate base is KC₂H₃O₂,
pKa = - log Ka = - log (1.8 x 10⁻⁵) = 4.74
so pH = 4.74 + log (0.2/0.2) = 4.74
This is called maximum buffer capacity (when acid conc. and base conc. are equal) the pH = pKa in this case
Acid is HC₂H₃O₂ and conjugate base is KC₂H₃O₂,
pKa = - log Ka = - log (1.8 x 10⁻⁵) = 4.74
so pH = 4.74 + log (0.2/0.2) = 4.74
This is called maximum buffer capacity (when acid conc. and base conc. are equal) the pH = pKa in this case
Explanation:
The given reaction at cathode will be as follows.
At cathode: ,
= -0.761 V
At anode: ,
= 0.761
Therefore, net reaction equation will be as follows.
Initial: 0.129 - - 0.427
Change: -0.047 - - -0.047
Equilibrium: (0.129 - 0.047) (0.427 - 0.047)
= 0.082 = 0.38
As for the given reaction is zero.
Hence, equation for calculating new cell potential will be as follows.
E_{cell} =
=
= 0.019
Thus, we can conclude that the cell potential of the given cell is 0.019.