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An unknown salt is either NaF, NaCl, or NaOCl. When 0.050 mol of the salt is dissolved in water to form 0.500 L of solution, the

victus00 [196]

Answer: The unknown salt is NaF

Explanation:

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

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

Moles of salt = 0.050 moles

Volume of solution = 0.500 L

Putting values in above equation, we get:

$\text{Molarity of salt}=\frac{0.050mol}{0.500L}\\\\\text{Molarity of salt}=0.1M$

To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:

pH + pOH = 14

We are given:

pH = 8.08

$pOH=14-8.08=5.92$

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

$pOH=-\log[OH^-]$

Putting values in above equation, we get:

$5.92=-\log[OH^-]$

$[OH^-]=10^{-5.92}=1.202\times 10^{-6}M$

The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH

The chemical equation for the hydrolysis of $X^-$ ions follows:

$X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b$

Initial: 0.1

At eqllm: 0.1-x x x

Concentration of $OH^-=x=1.202\times 10^{-6}M$

The expression of $K_b$ for above equation follows:

$K_b=\frac{[OH^-][HX]}{[X^-]}$

Putting values in above expression, we get:

$K_b=\frac{(1.202\times 10^{-6})\times (1.202\times 10^{-6})}{(1-(1.202\times 10^{-6}))}\\\\K_b=1.445\times 10^{-11}M$

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 = $1.445\times 10^{-11}$

Putting values in above equation, we get:

$10^{-14}=1.445\times 10^{-11}\times K_a\\\\K_a=\frac{10^{-14}}{1.445\times 10^{-11}}=6.92\times 10^{-4}$

We know that:

$K_a\text{ for HF}=6.8\times 10^{-6}$

$K_a\text{ for HCl}=1.3\times 10^{6}$

$K_a\text{ for HClO}=3.0\times 10^{-8}$

So, the calculated $K_a$ is approximately equal to the $K_a$ of HF

Hence, the unknown salt is NaF

6 0

3 years ago

Cómo se forma el enlace ionico

scoundrel [369]

Answer:

El enlace iónico se forma por transferencia de electrones entre metales y no metales. El enlace covalente se forma al compartir electrones entre no metales.

Explanation:

5 0

2 years ago

The less mass in a given volume of air the

DiKsa [7]

Answer:

The less mass in a given volume of air the less dense the air is going to be.

7 0

3 years ago

Read 2 more answers

A valid lewis structure of ________ cannot be drawn without violating the octet rule.

luda_lava [24]

A valid lewis structure of SO2 cannot be drawn without violating the octet rule.

correct me if i’m wrong.

7 0

2 years ago

If 1. 3618 moles of AsF3 are allowed to react with 1. 0000 mole of C2Cl6, what would be the theoretical yield of AsCl3, in moles

VARVARA [1.3K]

Answer:

AsF3:C2CI6

4:3

1.3618 moles: 1.02135 moles(1.3618÷4×3)

C2CI6 is the limting reagent

So the number of moles for AsCI3 is 0.817 moles( number of moles of the limting reagant) ÷3 ×4 (according to ratio by balancing chemical equation)=1.09 moles(3 s.f.)

or

Balanced equation

4AsF3 + 3C2Cl6 → 4AsCl3 + 3C2Cl2F4

Use stoichiometry to calculate the moles of AsCl3 that can be produced by each reactant.

Multiply the moles of each reactant by the mole ratio between it and AsCl3 in the balanced equation, so that the moles of the reactant cancel, leaving moles of AsCl3.

Explanation:

5 0

2 years ago