Question #36759

A volume of 80.0mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 14.2mL of 1.50 M H2SO4 was needed? The equation is
2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)

Expert's answer

M1*V1=M2*V2;

One mole of sulfuric acid neutralizes two moles of potassium hydroxide. Therefore:

M1=(14.2*1.5*2)/80=0.532M

One mole of sulfuric acid neutralizes two moles of potassium hydroxide. Therefore:

M1=(14.2*1.5*2)/80=0.532M

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