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3 years ago

What happens when Potassium Hydroxide reacts with Sulphuric Acid?

Chemistry

2 answers:

Ede4ka [16]3 years ago

5 0

As we know that neutralization reaction is a reaction in which base react with acid to form salt and water.

When Potassium Hydroxide reacts with Sulphuric Acid, it forms Potassium Sulphate and Water.

As a result of neutralization reaction, Potassium Sulphate and Water is formed.

2KOH + H2SO4 ----> K2SO4 + 2H2O

Here, K2SO4 is found in aqueous medium in neutralization reaction. It is a neutral salt.

4vir4ik [10]3 years ago

3 0

Here is your answer

We know that when an acid reacts with base, salt and water is formed.

This reaction is called Neutralisation reaction.

So,

2KOH + H2SO4 -> K2SO4(aq) + 2H2O

Reaction of Potassium Hydroxide with Sulphuric acid gives Potassium sulphate and water as product.

HOPE IT IS USEFUL

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Consider the following reaction: A(g)⇌2B(g). Find the equilibrium partial pressures of A and B for each of the following differe

Illusion [34]

Answer:

a. Kp=1.4

$P_{A}=0.2215 atm$

$P_{B}=0.556 atm$

b.Kp=2.0 * 10^-4

$P_{A}=0.495atm$

$P_{B}=0.00995 atm$

c.Kp=2.0 * 10^5

$P_{A}=5*10^{-6}atm$

$P_{B}=0.9999 atm$

Explanation:

For the reaction

A(g)⇌2B(g)

Kp is defined as:

$Kp=\frac{(P_{B})^{2}}{P_{A}}$

The conditions in the system are:

A B

initial 0 1 atm

equilibrium x 1atm-2x

At the beginning, we don’t have any A in the system, so B starts to react to produce A until the system reaches the equilibrium producing x amount of A. From the stoichiometric relationship in the reaction we get that to produce x amount of A we need to 2x amount of B so in the equilibrium we will have 1 atm – 2x of B, as it is showed in the table.

Replacing these values in the expression for Kp we get:

$Kp=\frac{(1-2x)^{2}}{x}$

Working with this equation:

$x*Kp=(1-2x)^{2} - -> x*Kp=4x^{2}-4x+1- - >4x^{2}-(4+Kp)*x+1=0$

This last expression is quadratic expression with a=4, b=-(4+Kp) and c=1

The general expression to solve these kinds of equations is:

$x=\frac{-b(+-)*\sqrt{(b^{2}-4ac)}}{2a}$ (equation 1)

We just take the positive values from the solution since negative partial pressures don´t make physical sense.

Kp = 1.4

$x_{1}=\frac{(1.4+4)+\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=1.128$

$x_{1}=\frac{(1.4+4)-\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=0.2215$

With x1 we get a partial pressure of:

$P_{A}=1.128 atm$

$P_{B}=1-2*1.128 = -1.256 atm$

Since negative partial pressure don´t make physical sense x1 is not the solution for the system.

With x2 we get:

$P_{A}=0.2215 atm$

$P_{B}=1-2*0.2215 = 0.556 atm$

These partial pressures make sense so x2 is the solution for the equation.

We follow the same analysis for the other values of Kp.

Kp=2*10^-4

X1=0.505

X2=0.495

With x1

$P_{A}=0.505atm$

$P_{B}=1-2*0.505 = -0.01005 atm$

Not sense.

With x2

$P_{A}=0.495atm$

$P_{B}=1-2*0.495 = 0.00995 atm$

X2 is the solution for this equation.

Kp=2*10^5

X1=50001

$X2=5*10^{-6}$

With x1

$P_{A}=50001atm$

$P_{B}=1-2*50001=-100001atm$

Not sense.

With x2

$P_{A}= 5*10^{-6}atm$

$P_{B}=1-2*5*10^{-6}= 0.9999 atm$

X2 is the solution for this equation.

8 0

3 years ago

Give the expression for the solubility product constant for baf2.

svetoff [14.1K]

Solubility product constant (Ksp) is applied to the saturated ionic solutions which are in equilibrium with its solid form. The solid is partially dissociated into its ions.

For the BaF, the dissociation as follows;
BaF₂(s) ⇄ Ba²⁺(aq) + 2F⁻(aq)

Hence,
Ksp = [Ba²⁺(aq)] [F⁻(aq)]²

3 0

3 years ago

What happens when Potassium Hydroxide reacts with Sulphuric Acid?​

What happens when Potassium Hydroxide reacts with Sulphuric Acid?