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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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barnuts

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

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