How many ML of 1.2 52 M KOH would be required to completely neutralize 9.55 ML of 0.114 HF

0.871 mL

Explanation:

Step 1: Write the balanced neutralization reaction

KOH + HF ⇒ KF + H₂O

Step 2: Calculate the reacting moles of HF

9.55 mL of 0.114 M HF react.

9.55 × 10⁻³ L × 0.114 mol/L = 1.09 × 10⁻³ mol

Step 3: Calculate the moles of KOH needed to react with 1.09 × 10⁻³ moles of HF

The molar ratio of KOH to HF is 1:1. The moles of KOH needed are 1/1 × 1.09 × 10⁻³ mol = 1.09 × 10⁻³ mol.

Step 4: Calculate the volume of KOH solution that contains 1.09 × 10⁻³ moles of KOH

An unknown volume of 1.252 M KOH solution contains 1.09 × 10⁻³ moles of KOH.

1.09 × 10⁻³ mol × 1 L/1.252 mol = 8.71 × 10⁻⁴ L = 0.871 mL

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.