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

colorless gas, exhibiting a blue glow when placed in an electric field

b) solid

its a type of rock