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Answer 1 · 2017-11-15T12:02:24

$"1.584 M"$

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 · 2017-11-15T13:22:55

Another approach ...

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