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2 years ago

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How should the ph of a 0.1 m solution of nac2h3o2 compare with that of a 0.1 m solution of kc2h3o2?

1 answer:

Semenov [28]2 years ago

7 0

We have that for the Question it can be said that the NaOH combines with CH_3COOH to produce CH_3COONa (Salt)

From the question we are told

how should the ph of a 0.1 m solution of <em>nac2h3o2</em> compare with that of a 0.1 m <u>solution </u>of kc2h3o2?

Generally

with the ph of a 0.1 m solution of <em>nac2h3o2</em> compared with that of a 0.1 m <u>solution </u>of kc2h3o2 ,we see that the salt produce is a weak acid and strong akali salt

We see that the salt produced in water gives a base from the derived weak acid

The salt produce is CH_3COONa

Therefore

the NaOH combines with CH_3COOH to produce CH_3COONa (Salt)

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Zinc dissolves in hydrochloric acid to yield hydrogen gas: Zn(s) + 2HCl(aq) --> ZnCl2(aq) + H2(g) When a 12.7 g chunk of zinc

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Answer:

$\boxed{\text{0.673 mol/L}}$

Explanation:

We are given the amounts of two reactants, so this is a limiting reactant problem.

1. Data:

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.

M_r: 65.38

Zn + 2HCl ⟶ ZnCl₂ + H₂

m/g: 12.7

V/mL: 5.00×10²

c/mol·L⁻¹: 1.450

2. Moles of each reactant

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$n = \text{12.7 g Zn} \times \dfrac{\text{1 mol Zn}}{\text{65.38 g Zn}} = \text{0.1942 mol Zn}$

(b) Moles of HCl

V = 5.0× 10² mL = 0.5000 L

$n = \text{0.5000 L HCl}\times \dfrac{\text{1.450 mol HCl}}{\text{1 L HCl}} = \text{0.7250 mol HCl}$

3. Identify the limiting reactant

Calculate the moles of ZnCl₂ obtained from each reactant

(i) From Zn

The molar ratio is 1 mol ZnCl₂:1 mol Zn

$n = \text{0.1942 mol Zn} \times \dfrac{\text{1 mol ZnCl}_{2}}{\text{1 mol Zn}} = \text{0.1942 mol ZnCl}_{2}$

(ii) From HCl

The molar ratio is 1 mol ZnCl₂:2 mol HCl

$n = \text{0.7250 mol HCl} \times \dfrac{\text{1 mol ZnCl}_{2}}{\text{2 mol HCl}} = \text{0.3625 mol ZnCl}_{2}$

Zinc is the limiting reactant, because it produces fewer moles of ZnCl₂.

4. Moles of HCl reacted

The molar ratio is 2 mol HCl:1 mol Zn

$n = \text{0.1942 mol Zn} \times \dfrac{\text{2 mol HCl}}{\text{1 mol Zn}} = \boxed{\text{0.3885 mol HCl}}$

5. Moles of HCl remaining

n = 0.7250 - 0.3885 = 0.3365 mol HCl

6. Concentration of hydrogen ions

The HCl is completely dissociated.

$c = \dfrac{\text{0.3365 mol}}{\text{0.5000 L}} = \textbf{0.673 mol/L}\\\\\text{The concentration of hydrogen ions is $\boxed{\textbf{0.673 mol/L}}$}$

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