Calculate normality and tire of KOH solution , if CN H2 C2 O4 solution , that is nessesary for titration is 0.1050 mol/1 and the Volume is 5.0 ml ; the volume of KOH - solution is 4.5 ml

Normality of KOH = 0.1062 N

Note: The given question is not clearly stated. A related question below is answered and explained;

What is the NORMALITY of a solution of KOH if 45.18 mL are required to neutralize 0.3g of pure oxalic acid H2C2O4 . 2H2O?

Explanation:

Equation of the reaction: 2KOH + H₂C₂O₄.2H₂O --> K₂C₂O₄ + 4H₂O

molar mass 0f KOH = 56 g/mol; molar mass of H₂C₂O₄.2H₂O = 126 g/mol

Number of moles of in 0.3 g of H₂C₂O₄.2H₂O = mass/molar mass

number of moles = 0.3 g / 126 g/mol = 0.0024 moles

From the equation of reaction, 2 moles of KOH are required to neutralize 1 mole of H₂C₂O₄.2H₂O

Number of moles of KOH required to neutralize 0.0024 moles of H₂C₂O₄.2H₂O = 2 * 0.0024 = 0.0048 moles

Molar concentration of 45.18 mL solution of KOH containing 0.0048 moles of  KOH = number of moles/volume (L)

molarity of KOH solution = (0.0048/45.18 mL) * (1000 mL/1 L) = 0.1062 mol/L

Normality = n * molarity

where is the number of moles of OH⁻ ions produces from  mole of KOH = 1

Normality of KOH = 1 * 0.1062

Normality of KOH = 0.1062 N

im not exactly sure but i think its b!

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