1.75 m hydrochloric acid is used in a titration with 47.8ml of sodium hydroxide, at the equivalence points 29.3ml of acid was added. how many moles of acid (h+) have been neutralized in this titration?

0.0512 mol

Explanation:

Molarity of a substance , is the number of moles present in a liter of solution .

M = n / V

M = molarity  

V = volume of solution in liter ,

n = moles of solute ,

From the question ,

Moles of hydrochloric acid is to be calculated ,

Molarity of hydrochloric acid = 1.75 M

Volume of hydrochloric acid = 29.3 ml

1 ml = 1 / 1000 L

Volume of hydrochloric acid = 29.3 ml /1000 = 0.0293 L

Using the above formula to calculate moles -

M = n / V

n = M * V

n = 1.75 M * 0.0293 L = 0.0512 mol

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