## Determine the [oh-] concentration of a 0.741 m koh solution at 25°c.

Question

Determine the [oh-] concentration of a 0.741 m koh solution at 25°c.

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1. KOH ionizes as follow,

KOHu2192 Ku207a + OHu207b

Concentration of OHu207b is0.741 M, calculating for pOH,

pOH = -log [OHu207b]

pOH = -log (0.741)

pOH = 0.130
Also,
pH + pOH = 14

pH + 0.130 = 14

pH = 14 – 0.130

pH = 13.87

2. The concentration of [tex]left[{{text{O}}{{text{H}}^-}}right][/tex] in KOH solution is [tex]boxed{{text{0}}{text{.741M}}}[/tex].

Further explanation:

Stoichiometry of a reaction is used to determine the amount of species present in the reaction by the relationship between the reactants and products. It can be used to determine the moles of a chemical species when the moles of other chemical species present in the reaction is given.

Consider the general reaction,

[tex]{text{A}}+2{text{B}}to3{text{C}}[/tex]

Here,

A and B are reactants.

C is the product.

One mole of A reacts with two moles of B to produce three moles of C. Stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3, and the stoichiometric ratio between B and C is 2:3.

Ionic compound:

Ionic compounds are the compounds that are formed from the ions of the respective species. Ions are the species that are formed either due to loss or gain of electrons. A neutral atom forms cation by the loss of electrons and anion by the gain of electrons.

Following are some of the properties of ionic compounds:

1. These are hard solids.

2. High melting and boiling points.

3. Good conductors of heat and electricity.

4. High enthalpy of fusion.

Ionic reaction:

Ionic reaction is a chemical reaction in which molecules dissociate into ions in aqueous solution. In ionic reaction net charge is same on both sides.

The equation for dissociation of [tex]{{text{H}}_2}{text{O}}[/tex]into ions is as follows:

[tex]{{text{H}}_2}{text{O}}to{{text{H}}^+}+{text{O}}{{text{H}}^-}[/tex]

Molarity

Molarity is the concentration term defined as the number of moles of solute particles present per litre of solution. It is denoted by M, and the unit of molarity is mol/L.

Mathematically, it is written as follows:

[tex]{text{Molarity(M)}}=frac{{{text{number of moles of solute}}}}{{{text{volume of solution}}}}[/tex]

Given the concentration of KOH is 0.741 M. This means that 0.741 moles of KOH are present in 1 litre of solution.

The reaction for dissociation of KOH into ions is as follows:

[tex]{text{KOH}}to{{text{K}}^+}+{text{O}}{{text{H}}^-}[/tex] …… (1)

According to stoichiometry of reaction in equation (1), 1 mole of KOH dissociates to form 1 mole of [tex]{{text{K}}^+}[/tex]and 1 mole of [tex]{text{O}}{{text{H}}^-}[/tex].

Therefore 0.741 mol of KOH dissociates to forms 0.741 mol of [tex]{{text{K}}^+}[/tex]and 0.741 mol of [tex]{text{O}}{{text{H}}^-}[/tex]in solution. So the concentration of [tex]{text{O}}{{text{H}}^-}[/tex]and [tex]{{text{K}}^+}[/tex]ions in solution is equal to concentration to KOH that is 0.741 M.

Hence the concentration of [tex]left[ {{mathbf{O}}{{mathbf{H}}^{mathbf{-}}}}right][/tex] is 0.741 M.

1. DrawLewis structure of ionic compoundhttps://brainly.com/question/6786947.

2. Identify the neutral element https://brainly.com/question/9616334