Question

How many grams of 4.15% (w/w) solution of KOH in water is needed to neutralize completely the acid in 13.1 mL of 0.385 M H2SO4?

Answer #1

Reaction of H2SO4 with KOH

H2SO4 + 2 KOH -------> K2SO4 + 2 H2O

Concentration of H2SO4 = 0.385 M = 0.385 mol/L

Volume of H2SO4 solution = 13.1 ml = 13.1 L / 1000 =0.0131 L

Number of moles of H2SO4 = 0.385 mol/L * 0.0131 L = 0.00504 mol

Frome reaction, for complete neutralization

Number of moles of KOH = 2 * Number of moles of H2SO4

Number of moles of KOH = 2 * 0.00504 mol = 0.01008 mol

Molar mass of KOH = 56.1 g/mol

Mass of pure KOH needed = 56.1 g/mol * 0.01008 mol = 0.5655 g

Mass of 4.15%(w/w) KOH solution needed = 0.5655 g / 0.0415 = 13.63 g --------Ans

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_________ mL
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