K2S(aq) + Co(NO3)2(aq) → CoS (?) + KNO3 (?)

Your job is to determine the concentration of ammonia in a commercial window cleaner. In the titration of a 25.0 mL sample of th

ruslelena [56]

Answer:

The initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of HCl solution = 0.164 M

Volume of solution = 23.8 mL = 0.0238 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.164M=\frac{\text{Moles of HCl}}{0.0238L}\\\\\text{Moles of HCl}=(0.146mol/L\times 0.0238L)=0.0035mol$

The chemical equation for the reaction of ammonia and HCl follows:

$NH_3+HCl\rightarrow NH_4^++Cl^-$

By Stoichiometry of the reaction:

1 mole of HCl reacts with 1 mole of ammonia

So, 0.0035 moles of HCl will react with = $\frac{1}{1}\times 0.0035=0.0035mol$ of ammonia

Calculating the initial concentration of ammonia by using equation 1:

Moles of ammonia = 0.0035 moles

Volume of solution = 25 mL = 0.025 L

Putting values in equation 1, we get:

$\text{Initial concentration of ammonia}=\frac{0.0035mol}{0.025L}=0.14M$

By Stoichiometry of the reaction:

1 mole of ammonia produces 1 mole of ammonium ion

So, 0.0035 moles of ammonia will react with = $\frac{1}{1}\times 0.0035=0.0035mol$ of ammonium ion

Calculating the concentration of ammonium ion by using equation 1:

Moles of ammonium ion = 0.0035 moles

Volume of solution = [23.8 + 25] mL = 48.8 mL = 0.0488 L

Putting values in equation 1, we get:

$\text{Molarity of ammonium ion}=\frac{0.0035mol}{0.0488L}=0.072M$

To calculate the acid dissociation constant for the given base 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

$K_b$ = Base dissociation constant = $1.8\times 10^{-5}$

$10^{-14}=1.8\times 10^{-5}\times K_a\\\\K_a=\frac{10^{-14}}{1.8\times 10^{-5}}=5.55\times 10^{-10}$

The chemical equation for the dissociation of ammonium ion follows:

$NH_4^+\rightarrow NH_3+H^+$

The expression of $K_a$ for above equation follows:

$K_a=\frac{[NH_3][H^+]}{[NH_4^+]}$

We know that:

$[NH_3]=[H^+]=x$

$[NH_4^+]=0.072M$

Putting values in above expression, we get:

$5.55\times 10^{-10}=\frac{x\times x}{0.072}\\\\x=6.32\times 10^{-6}M$

To calculate the pH concentration, we use the equation:

$pH=-\log[H^+]$

We are given:

$[H^+]=6.32\times 10^{--6}M$

$pH=-\log (6.32\times 10^{-6})\\\\pH=5.20$

Hence, the initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20

5 0

3 years ago

I got c, just wondering was it right?

Talja [164]

Answer:

Yes you are right.

Explanation:

4 0

2 years ago

Read 2 more answers

What is responsible for the blue sky?

andreev551 [17]

Answer:

well for me it's

Explanation:

The reflection of the seas

7 0

3 years ago

Read 2 more answers

In ionic compounds, ________ lose their valence electrons to form positively charged ________. nonmetals, anions metals, polyato

Keith_Richards [23]

Answer:
In ionic compounds, Metals lose their valence electrons to form positively charged Cations.

Explanation:
Metals have the ability to loose elctrons readily. For example metals of Group IA and Group IIA readily looses electrons in order to obtain Noble Gas Configuration. On the other hand Non-metals tends to gain electrons and acquire negative charge. While Ions are made when an an element gain or loose electrons. After loosing electrons element get positive charge which is called as Cation while on gaining electron it gets negative charge called as Anion.

6 0

3 years ago

Read 2 more answers

7.895 g + 3.4 g round to the correct number significant figures

DIA [1.3K]

Answer:

11.3 g

Explanation:

7.895 + 3.4 = 11.295

When rounded to correct number of significant figures --> 11.3

There are 3 significant figures in 11.3

7 0

3 years ago