An aqueous solution in a 55 gallon (208 l drum), characterized by minimal buffering capacity, received 4kg of phenol and 1.5 kg of sodium phenate. What is the ph of the solution. The pka of phenol = 9.98. Mw of phenol and sodium phenate are 94 g/mol and 116 g/mol, respectively.
Volume of solution = 55 gallons = 208.2 L [ 1 gallon = 3.78 L]
moles of phenol = mass / molar mass = 4000 g / 94 = 42.55 moles
moles of sodium phenate = mass / molar mass = 1500 / 116 = 12.93 moles
pKa of phenol = 9.98
We know that the pH of buffer is calculated using Hendersen Hassalbalch's equation
pH = pKa + log [salt] / [acid]
volume is same for both the sodium phenate and phenol has we can directly take the moles of each in the formula
pH = 9.98 + log [12.93 / 42.55] = 9.46
I think it is decomposition because it processes in which chemical species break up into simpler parts.
Answer:
0.45 moles
Explanation:
The computation of the number of moles left in the cylinder is shown below:
As we know that
we can say that
where,
n1 = 1.80 moles of gas
V2 = 12.0 L
And, the V1 = 48.0 L
Now placing these values to the above formula
So, the moles of gas in n2 left is
= 0.45 moles
We simply applied the above formulas so that the n2 moles of gas could arrive
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