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Answer:
7.37 mL of KOH
Explanation:
So here we have the following chemical formula ( already balanced ), as HNO3 reacts with KOH to form the products KNO3 and H2O. As you can tell, this is a double replacement reaction,
HNO3 + KOH → KNO3 + H2O
Step 1 : The moles of HNO3 here can be calculated through the given molar mass ( 0.140 M HNO3 ) and the mL of this nitric acid. Of course the molar mass is given by mol / L, so we would have to convert mL to L.
Mol of NHO3 = 0.140 M 30 / 1000 L = 0.140 M
0.03 L = .0042 mol
Step 2 : We can now convert the moles of HNO3 to moles of KOH through dimensional analysis,
0.0042 mol HNO2 ( 1 mol KOH / 1 mol HNO2 ) = 0.0042 mol KOH
From the formula we can see that there is 1 mole of KOH present per 1 moles of HNO2, in a 1 : 1 ratio. As expected the number of moles of each should be the same,
Step 3 : Now we can calculate the volume of KOH knowing it's moles, and molar mass ( 0.570 M ).
Volume of KOH = 0.0042 mol ( 1 L / 0.570 mol )
( 1000 mL / 1 L ) = 7.37 mL of KOH
Answer : The concentration of a solution with an absorbance of 0.420 is, 0.162 M
Explanation :
Using Beer-Lambert's law :
As per question, at constant path-length there is a direct relation between absorbance and concentration.
where,
A = absorbance of solution
C = concentration of solution
l = path length
= initial absorbance = 0.350
= final absorbance = 0.420
= initial concentration = 0.135 M
= final concentration = ?
Now put all the given value in the above relation, we get:
Thus, the concentration of a solution with an absorbance of 0.420 is, 0.162 M