When preparing the diazonium salt, the solution is tested with potassium iodide-starch paper. a positive test is the immediate f
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Answer:
Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.
Explanation:
Volume of NaOH = 1.7 ml = 0.0017 L
Molarity of NaOH = 0.0811 M
Moles of NaOH = n
n = 0.0001378 mol
According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.
Then 0.0001378 mol of NaOH will neutralize:
of sulfuric acid.
Concentration of sulfuric acid in the acid rain sample: x
Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.
[H_{3}O^{+}] = 0.00770 M
The equilibrium equation representing the dissociation of
Given [H_{3}O^{+}] = 0.00770 M
Let the initial concentration of acid be x and change y
So y = =
= 0.00770 M
0.00257 x - 0.00001979 = 0.00005929
x = 0.031 M
Therefore, initial concentration of the weak acid is <u>0.031 M</u>