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Varvara68 [4.7K]
3 years ago
8

Fill in the left side of this equilibrium constant equation for the reaction of nitrous acid (HNO_2) with water.

Chemistry
2 answers:
koban [17]3 years ago
6 0

Answer:

1. The equilibrium equation for the reaction between nitrous acid (HNO2) and water (H2O) is given below:

HNO2(aq) + H2O(l) <=> H3O+(aq) + NO2- (aq)

2. Ka = [H3O+] [NO2-] / [HNO2]

Explanation:

1. Step 1:

The reaction of nitrous acid HNO2 acid with water.

When nitrous acid react with water, it will produce hydronium ion (H3O+) and the corresponding nitrite ion (NO2-). This is so because an acid will dissolves in water to produce hydronium ion as the only positive ion. The equation is illustrated below:

HNO2(aq) + H2O(l) <=> H3O+(aq) + NO2- (aq)

2. Step 2:

Writing the Ka for the above equation. This is illustrated below:

The Ka is given by:

Ka = [H3O+] [NO2-] / [HNO2]

mafiozo [28]3 years ago
5 0

Answer:

Ka=\frac{[H^+][NO_2^-]}{[HNO_2]}

Explanation:

The reaction

The reaction of nitrous acid with water is a dissociation reaction of an acid:

HNO_2(aq) + H_2O(l) \longrightarrow H^+(aq) + NO_2^-(aq) + H_2O(l)

HNO_2(aq) \longrightarrow H^+(aq) + NO_2^-(aq)

The Ka equation

Ka=\frac{[H^+][NO_2^-]}{[HNO_2]}

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Consider the titration of 1L of 0.36 M NH3 (Kb=1.8x10−5) with 0.74 M HCl. What is the pH at the equivalence point of the titrati
worty [1.4K]

Answer:

C

Explanation:

The question asks to calculate the pH at equivalence point of the titration between ammonia and hydrochloric acid

Firstly, we write the equation of reaction between ammonia and hydrochloric acid.

NH3(aq)+HCl(aq)→NH4Cl(aq)

Ionically:

HCl + NH3 ---> NH4  +  Cl-

Firstly, we calculate the number of moles of  the ammonia  as follows:

from c = n/v and thus, n = cv = 0.36 × 1 = 0.36 moles

At the equivalence point, there is equal number of moles of ammonia and HCl.

Hence, volume of HCl = number of moles/molarity of HCl = 0.36/0.74 = 0.486L

Hence, the total volume of solution will be 1 + 0.486 = 1.486L

Now, we calculate the concentration of the ammonium ions = 0.36/1.486 = 0.242M

An ICE TABLE IS USED TO FIND THE CONCENTRATION OF THE HYDROXONIUM ION(H3O+). ICE STANDS FOR INITIAL, CHANGE AND EQUILIBRIUM.

                 NH4+      H2O     ⇄  NH3        H3O+

I                0.242                           0             0

C                 -X                              +x              +X

E             0.242-X                          X              X

Since the question provides us with the base dissociation constant value K b, we can calculate the acid dissociation constant value Ka

To find this, we use the mathematical equation below

K a ⋅ K b    = K w

 

, where  K w- the self-ionization constant of water, equal to  

10 ^-14  at room temperature

This means that you have

K a = K w.K b   = 10 ^− 14 /1.8 * 10^-5 =  5.56 * 10^-10

Ka = [NH3][H3O+]/[NH4+]

= x * x/(0.242-x)

Since the value of Ka is small, we can say that 0.242-x ≈  0.242

Hence, K a = x^2/0.242 = 5.56 * 10^-10

x^2 = 0.242 * 5.56 * 10^-10 = 1.35 * 10^-10

x = 0.00001161895

[H3O+] = 0.00001161895

pH = -log[H3O+]

pH = -log[0.00001161895 ] = 4.94

7 0
3 years ago
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