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AVprozaik [17]
3 years ago
10

Does the reaction of a main-group metal oxide in water produce an acidic solution or a basic solution? Write a balanced equation

for the reaction of a Group 2A(2) oxide with water.
Chemistry
1 answer:
Fynjy0 [20]3 years ago
4 0

<u>Answer:</u> The main group metal produce a basic solution in water and the reaction is MO+H_2O\rightarrow M(OH)_2

<u>Explanation:</u>

Main group elements are the elements that are present in s-block and p-block.

The metals that are the main group elements are located in Group IA, Group II A and Group III A.

Oxides are formed when a metal or a non-metal reacts with oxygen molecule. There are two types of oxides which are formed: Acidic oxides and basic oxides.

  • Acidic oxides are formed by the non-metals.
  • Basic oxides are formed by the metals.

When a metal oxide is reacted with water, it leads to the formation of a base.

The general formula of the oxide formed by Group II-A metals is 'MO'

The chemical equation for the reaction of metal oxide of Group II-A and water follows:

MO+H_2O\rightarrow M(OH)_2

Hence, the main group metal produce a basic solution in water and the reaction is MO+H_2O\rightarrow M(OH)_2

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3 years ago
Calculate the ph at of a solution of sodium hypochlorite . note that hypochlorous acid is a weak acid with a of . round your ans
loris [4]

The question is incomplete, here is the complete question:

Calculate the pH at 25°C of a 0.39 M solution of sodium hypochlorite NaClO. Note that hypochlorous acid HClO is a weak acid with a pKa of 7.50. Round your answer to 1 decimal place.

<u>Answer:</u> The pH of the solution is 10.4

<u>Explanation:</u>

We are given:

Molarity of sodium hypochlorite = 0.39 M

pK_a of HClO = 7.50

We know that:

pK_a=-\log K_a

K_a  of HClO = 10^{-7.50}=3.16\times 10^{-8}

To calculate the base dissociation constant for the given acid dissociation constant, we use the equation:

K_w=K_b\times K_a

where,

K_w = Ionic product of water = 10^{-14}

K_a = Acid dissociation constant  = 3.16\times 10^{-8}

K_b = Base dissociation constant

Putting values in above equation, we get:

10^{-14}=3.16\times 10^{-8}\times K_b\\\\K_b=\frac{10^{-14}}{3.16\times 10^{-8}}=3.16\times 10^{-7}

The chemical equation for the reaction of hypochlorite ion with water follows:

                    ClO^-+H_2O\rightarrow HClO+OH^-

<u>Initial:</u>           0.39

<u>At eqllm:</u>      0.39-x                   x           x

The expression of K_b for above equation follows:

K_b=\frac{[HClO][OH^-]}{[ClO^-]}

Putting values in above equation, we get:

3.16\times 10^{-7}=\frac{x\times x}{(0.39-x)}\\\\x=-0.00035,0.00035

Neglecting the negative value of 'x' because concentration cannot be negative

To calculate the pOH of the solution, we use the equation:

pOH=-\log[OH^-]

We are given:

[OH^-]=0.00035M

Putting values in above equation, we get:

pOH=-\log (0.00035)=3.6

To calculate pH of the solution, we use the equation:

pH+pOH=14\\pH=14-3.6=10.4

Hence, the pH of the solution is 10.4

3 0
3 years ago
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