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Drupady [299]
1 year ago
13

Write the chemical equation for the autoionization of water and label the conjugate

Chemistry
1 answer:
Vinvika [58]1 year ago
3 0

The equation for the autoionization of water is-

NH3 + H2O → NH4+ + OH− (H2O acts as an acid)

HCl + H2O → H3O+ + Cl− (H2O acts as a base)

<h3>What do you mean by the autoionization of water ?</h3>

Autoionization of water, the autoionization constant Kw, and the relationship between [H⁺] and [OH⁻] in aqueous solutions.

Water can undergo autoionization to form and ions.

The expression for the autoionization constant is

The expression for the autoionization constant isK_\text{w}=[\text{H}_3\text{O}^+][\text{OH}^-]\quad\quad\text{(Eq. 1)}K w​ =[H 3​ O + ][OH − ]

In this way the autoionization of water,

NH3 + H2O → NH4+ + OH− (H2O acts as an acid)

HCl + H2O → H3O+ + Cl− (H2O acts as a base) is given.

Learn more about autoionization process ,here:

brainly.com/question/14993418

#SPJ5

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We have to know the molarity of solution obtained when 5.71 g of Na₂CO₃.10 H₂O is dissolved in water and made up to 250 cm³ solution.

The molarity of solution obtained when 5.71 g of sodium carbonate-10-water (Na₂CO₃.10 H₂O)  is dissolved in water and made up to 250.0 cm^3 solutionis: (A) 0.08 mol dm⁻³

The molarit y of solution means the number of moles of solute present in one litre of solution. Here solute is Na₂CO₃.10 H₂O and solvent is water. Volume of solution is 250 cm³.

Molar mass of Na₂CO₃.10 H₂O is 286 grams which means mass of one mole of Na₂CO₃.10 H₂O is 286 grams.

5.71 grams of Na₂CO₃.10 H₂O is equal to \frac{5.71}{286}= 0.0199 moles of Na₂CO₃.10 H₂O. So, 0.0199 moles of Na₂CO₃.10 H₂O present in 250 cm³ volume of solution.

Hence, number of moles of Na₂CO₃.10 H₂O present in one litre (equal to 1000 cm³) of solution is \frac{0.0199 X 1000}{250} = 0.0796 moles. So, the molarity of the solution is 0.0796 mol/dm³ ≅ 0.08 mol/dm³


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