Question

Question #b2f7a

Answer 1

`"1.584 M"`

Explanation:

Formula of Molarity is

`"Molarity" = "Moles of solute"/"Volume of solution (in litres)"`

Moles of KOH `= "1.80 M × 17.6" cancel("ml") × (10^-3 "L")/cancel"ml" = "0.03168 mol"`

Acidity of KOH is 1. So, equivalents of KOH is 0.03168 mol/eq

For complete neutralisation equivalents of KOH should be equal to equivalents of `H_2SO_4`

Moles of `H_2SO_4` `= "Equivalents"/2 = "0.03168 mol"/2 = "0.01584 mol"`

Molarity of `H_2SO_4` `= "0.01584 mol"/(10 cancel"ml" × 10^-3 "L"/cancel"ml") = "1.584 M"`

Answer 2

Another approach ...

Explanation:

Consider the metathesis reaction of the titration ...
`H_2SO_4(aq)` + `2NaOH(aq)` => `Na_2SO_4` + `2H_2O`
10 ml `H_2SO_4` + 17.6ml(1.80M KOH)

use relationship ...
2(Molarity of Acid Soln x Volume of Acid Soln) = (Molarity of Base Soln x Volume of Base Soln)

Substitute given data and solve for Molarity of Acid Soln
`"Molarity"_(acid)` = `((1.80M)(17.6ml))/(2(10.0ml))` = `1.584M` Solution