The chemical symbol for sodium is <br> a. na. <br> b. sn. <br> c. s. <br> d. n.

We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

5 0

3 years ago

an unknown molecule is found to consist of 24.2% carbon by mass, 4.0% hydrogen by mass and the remaining mass is due to chlorine

Paha777 [63]

Answer:

C3 H6 Cl 3

Explanation:

C -24.2%

H - 4.0%

Cl - (100-24.2 - 4.0)=73.8 %

We can take 100g of the substance, then we have

C -24.2 g

H - 4.0 g

Cl - 73.8 g

Find the moles of these elements

C -24.2 g/12.0 g/mol =2.0 mol

H - 4.0 g/1.0 g/mol = 4. 0 mol

Cl - 73.8 g/ 35.5 g/mol = 2.1 mol

Ratio of these elements gives simplest formula of the substance

C : H : Cl = 2 : 4 : 2 = 1 : 2 : 1

CH2Cl

Molar mass (CH2Cl) = 1*12.0 +2*1.0 + 1*35.5 = 49.5 g/mol

Real molar mass = 150 g/mol

real molar mass/ Molar mass (CH2Cl) = 150 /49.5=3

So, Real formula should be C3 H6 Cl 3.

4 0

3 years ago

Read 2 more answers

Calculate the numbers of mL in 0.603 L

Artist 52 [7]

Answer:

603 mL

Explanation:

A milliliter is a unit of volume equal to 1/1000th of a liter. It is the same as a cubic centimeter.

6 0

3 years ago

To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 2.2-L bulb, then filled

Ber [7]

Answer:

N2

Explanation:

We use the ideal gas equation to calculate the number of moles of the diatomic gas. Then from the number of moles we can get

Given:

P = 2atm

1atm = 101,325pa

2atm = 202,650pa

T = 27 degrees Celsius = 27 + 273.15 = 300.15K

V = 2.2L

R = molar gas constant = 8314.46 L.Pa/molK

PV = nRT

Rearranging n = PV/RT

Substituting these values will yield:

n = (202,650 * 2.2)/(8314.46* 300.15)

n = 0.18 moles

To get the molar mass, we simply divide the mass by the number of moles.

5.1/0.18 = 28.5g/mol

This is the closest to the molar mass of diatomic nitrogen N2.

Hence, the gas is nitrogen gas

7 0

3 years ago

How many moles of Au atoms are present in one ounce of pure gold? [1 oz = 28.4 g]

IgorLugansk [536]

Answer:

3.36x10^27 atoms

3 0

3 years ago

The chemical symbol for sodium is a. na. b. sn. c. s. d. n.