Question

Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory to standardize strong base solutions. It has the unwieldy formula of KHC₈H₄O₄. This is often written in shorthand notation as KHP.

How many grams of KHP are needed to exactly neutralize 21.6 mL of a 0.479 M potassium hydroxide solution ?

What volume of a 0.379 M potassium hydroxide solution is needed to exactly neutralize 2.09 grams of KHP ?

Answer 1

Ans - The reaction between potassium potassium hydrogen plethalate can hydroxide be written and as KOH + kH Ca HAO K ₂ Cg Hala + 60 This is a balanced equation from the balanced equation, I mole of kot reacts with Inde of KHCg Hala (KKP) Given concentation of kot solution = 0.479 m Varume of solution - 21.6ml = 0.0216 L We have litre molarity - no. of miles of solute (n) volume of the solution in 2) 0.979 m = ② 0.02162 2) n = 0.479 x 0. 0216 mois = 0.0103 moles. I mole of kon is neutralised by Imole of kHP 0.0103 moles of kon will be neutralised by 0.0103 molus of kHP mlar man of kHP = 204. 22 g/mol

We have no. of moles - man mdan man or man - no. of moles & molar mans = 0.0103 204.22 g = 2.10g 204.22 Grams of kHP needed - 2010g Gwen grams of kHP = 2.09g molar man of KHP = 204.22 gl mo .. no. of molus of KAP = 2.09 mars = 0.0102 moles from the balanced equation, I mole of KHP neutralised Imole of kot : 0.0102 moles of kHP will neutralised 0.0102 mols of Koh Given concentration of kot = 0.379 m. We have molarity = no. of moles of solute volume of solution in litre. 2) Volume = no. of mous molarity - 0.0102 L 0.379 = 0·0269 L volume needed = 26.9 ml