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A volume of 55.6 ml of aqueous potassium hydroxide is titrated against a standard solution of sulfuric acid. What was the molarity of the potassium hydroxide solution if 15.1 ml of 1.50 m sulfuric acid was needed

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  1. 19 May, 14:06
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    The molar concentration of the potassium hydroxide solution was 0.815 mol/L.

    Balanced equation: 2KOH + H_2SO_4 → K_2SO_4 + 2H_2O

    Moles of H_2SO_4:

    15.1 mL H_2SO_4 * (1.50 mmol H_2SO_4 / 1 mL H_2SO_4)

    = 22.65 mmol H_2SO_4

    Moles of KOH: 22.65 mmol H_2SO_4 * (2 mmol KOH/2 mmol H_2SO_4)

    = 45.30 mmol KOH

    Concentration of KOH: c = "moles"/"litres" = 45.30 mmol/55.6 mL

    = 0.815 mol/L
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