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statuscvo [17]
4 years ago
5

Write a balanced complete ionic equation for: hi(aq)+rboh(aq)→

Chemistry
2 answers:
neonofarm [45]4 years ago
6 0

The complete ionic equation for the reaction is as follows:

\boxed{{{\mathbf{H}}^ + }\left( q \right) + {{\mathbf{I}}^ - }\left( {aq} \right) + {\mathbf{R}}{{\mathbf{b}}^ + }\left( {aq} \right) + {\mathbf{O}}{{\mathbf{H}}^ - }\left( {aq} \right) \to {{\mathbf{H}}_{\mathbf{2}}}{\mathbf{O}}\left( l \right) + {{\mathbf{I}}^ - }\left( {aq} \right) + {\mathbf{R}}{{\mathbf{b}}^ + }\left( {aq} \right)}

Further Explanation:

Double displacement reaction is defined as the reaction in which ions of two compound interchange with each other to form the product. For example, the general double displacement reaction between two compounds AX and BY  is as follows:

{\text{AX}} + {\text{BY}} \to {\text{AY}} + {\text{BX}}

The three types of equations that are used to represent the chemical reaction are as follows:

1. Molecular equation

2. Complete ionic equation

3. Net ionic equation

The reactants and products remain in undissociated form in molecular equation. In the case of complete ionic equation, all the ions that are dissociated and present in the reaction mixture are represented while in the case of net ionic equation only the useful ions that participate in the reaction are represented.

The steps to write the complete ionic reaction are as follows:

Step 1: Write the molecular equation for the reaction with the phases in the bracket.

In the reaction, HI reacts with RbOH to form RbI and {{\text{H}}_{\text{2}}}{\text{O}}. The balanced molecular equation of the reaction is as follows:

 {\text{HI}}\left( {aq} \right) + {\text{RbOH}}\left( {aq} \right) \to {\text{RbI}}\left( {aq} \right){\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\left( l \right)

Step 2: Dissociate all the compounds with the aqueous phase to write the complete ionic equation. The compounds with solid and liquid phase remain same. The complete ionic equation is as follows:

{{\mathbf{H}}^ + }\left( q \right) + {{\mathbf{I}}^ - }\left( {aq} \right) + {\mathbf{R}}{{\mathbf{b}}^ + }\left( {aq} \right) + {\mathbf{O}}{{\mathbf{H}}^ - }\left( {aq} \right) \to {{\mathbf{H}}_{\mathbf{2}}}{\mathbf{O}}\left( l \right) + {{\mathbf{I}}^ - }\left( {aq} \right) + {\mathbf{R}}{{\mathbf{b}}^ + }\left( {aq} \right)

Learn more:

1. Balanced chemical equation brainly.com/question/1405182

2. Oxidation and reduction reaction brainly.com/question/2973661

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Chemical reaction and equation

Keywords: Double displacement reaction, types of equation, molecular equation, complete ionic equation, net ionic equation, RbI, RbOH, H2O, HI, chemical reaction.

mestny [16]4 years ago
3 0

The balanced complete ionic equation for HI(aq) + RbOH(aq) →RbI(aq) +H2O(l) is given by;

Net ionic equation; H+(aq) + OH-(aq) → H2O(l)

<h2>Further Explanation: </h2><h3>Balancing chemical equation: </h3>
  • Balancing chemical equations is a try and error method that ensures the number of atoms in the side of the reactants is equal to the number of atoms in the side of products.
  • Balancing chemical equations ensures that the law of conservation of mass is observed.
  • Balancing chemical equations may also require inclusion of state symbols which shows the state of each compound or element involved in the chemical reaction.
<h3>Net ionic equations  </h3>
  • Net ionic equations that do not include all ions that are present in the reactants and products.
  • The equations only includes the ions that are involved in the reaction.
  • The ions that are not involved in a chemical reaction are called spectator ions and are cancelled out while writing ionic equations.
  • For Example;

A reaction between HI and  RbOH is given by;

HI(aq) + RbOH(aq) →RbI(aq) + H2O(l)

  • The ionic equation for the reaction will be;

H+(aq) + I-(aq) + Rb+(aq) + OH-(aq) → H2O(l) + Rb+(aq) + I-(aq)

  • H+ changes state from aqueous to liquid, OH- changes state from aqueous to liquid, while both Rb+ and I- do not change state and thus are not included while writing the net ionic equation.

Therefore;

Net ionic equation will be;

H+(aq) + OH-(aq) → H2O(l)

Keywords: Net ionic equations

<h3>Learn more about;  </h3>
  • Balancing of chemical equations: brainly.com/question/5297242
  • Net ionic equation: brainly.com/question/5552670  
  • Spectator ions: brainly.com/question/5552670
  • Example of net ionic equations; brainly.com/question/5552670

Level: High school  

Subject: Chemistry  

Topic: Chemical equation  

Sub-topic: Net ionic equations

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sukhopar [10]

3.08 is the pOH of a 0.175 M aqueous solution of NX_3.

0.215% is the per cent ionization of a 0.325 M aqueous solution of NX_3

<h3>What is pH?</h3>

pH is a measure of how acidic/basic water is.

A)

NX_3 + H_2O →NHX_3^+ + OH^-

Kb = 4.5 x10^-6

Kb = {concentration of (NH₄⁺) x concentration of (OH⁻)} ÷ concentration of (NH₃).

concentration of (NH₄⁺) = concentration of (OH⁻) = x.

x² = Kb x concentration of (NH₃)

x² = 4.5 × 10⁻⁶ × 0.175 = 7.0 × 10⁻⁷.

x = concentration of (OH⁻) = √(7.0 × 10⁻⁷)

= 8.367 × 10⁻⁴

pOH = -log(c(OH⁻))

=- log ( 8.367 × 10⁻⁴)

= 3.08

B)

Chemical reaction: NX₃ + H₂O ⇄ NX₃H⁺ + OH⁻.

Concentration of (NX₃) = 0.325 M.

Kb = 4.5 x 10⁻⁶.

[NX₃H⁺] = [OH⁻] = x.

[NX₃] = 0.325 M - x.

Kb = [NX₃H⁺] x [OH⁻] ÷  [NX₃].

4.5 x 10⁻⁶ = x² ÷ (0.325 M - x).

x = 0.0007 M.

Per cent of ionization:

α = 0. 0007 M ÷ 0. 325 M x 100%

= 0.215%.

Hence,

3.08 is the pOH of a 0.175 M aqueous solution of NX_3.

0.215% is the per cent ionization of a 0.325 M aqueous solution of NX_3

Learn more about pH here:

brainly.com/question/12353627

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<h3>Further explanation</h3>

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