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

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.