A volume of 90.0 ml of aqueous potassium hydroxide (koh) was titrated against a standard solution of sulfuric acid (h2so4). what was the molarity of the koh solution if 15.7 ml of 1.50 m h2so4 was needed? the equation is 2koh (aq) + h2so4 (aq) →k2so4 (aq) + 2h2o (l)

Question

Tyrese Bradford

Chemistry

2 November, 19:21

A volume of 90.0 ml of aqueous potassium hydroxide (koh) was titrated against a standard solution of sulfuric acid (h2so4). what was the molarity of the koh solution if 15.7 ml of 1.50 m h2so4 was needed? the equation is 2koh (aq) + h2so4 (aq) →k2so4 (aq) + 2h2o (l)

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Emmalee Oconnor · Answer 1

From the equation we can see that 2 moles of KOH is required for every 1 mole of H2SO4.

So calculate the moles of H2SO4:

moles H2SO4 = 1.50 M * 0.0157 L = 0.02355 mol

So moles of KOH is:

moles KOH = 0.02355 mol * 2 = 0.0471 mol

The concentration in molarity is therefore:

Molarity = 0.0471 mol / 0.090 L = 0.52 M