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ExtremeBDS [4]
3 years ago
5

A volume of 30.0 mL of a 0.140 M HNO3 solution is titrated with 0.570 M KOH. Calculate the volume of KOH required to reach the e

quivalence point.
Chemistry
1 answer:
igomit [66]3 years ago
5 0

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

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1. The measurement turned out to be so expensive since the student did not rely on the significant figures to calculate the edge of the cube and approximate it to the next value (2.1 cm possibly) that would allow a simpler construction and therefore its cost was much lower .

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Since the figure to be constructed is a cube, he had to calculate the cube root of the volume to find the value of the edge of the cube:

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If you want to learn more about exercises with significant figures, you can see the next link: brainly.com/question/11904364?referrer=searchResults

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