All categories

3 years ago

Identify the major ionic species present in an aqueous solution of sodium sulfate.

Chemistry

2 answers:

KATRIN_1 [288]3 years ago

5 0

The major ionic species present in aqueous solution of sodium sulfate is

2 Na+ and SO4^2-

explanation

When electrolyte dissolve in water they dissociate into ions.

There are two type of electrolyte that is strong and weak electrolyte.

Sodium sulfate is a strong electrolyte and therefore it dissociate completely into ions as below

Na2So4 → 2 Na+ (aq) + So4^2- (aq)

Sonja [21]3 years ago

5 0

Answer: The major ionic species present in the solution are sodium and sulfate ions.

Explanation:

Sodium sulfate is an ionic compound because it is formed by the complete transfer of electrons from sodium ions to sulfate ions.

When an ionic compound is dissolved in water, it dissociates into its corresponding ions.

The chemical equation for the ionization of sodium sulfate follows:

$Na_2SO_4(aq.)\rightarrow 2Na^+(aq.)+SO_4^{2-}(aq.)$

By Stoichiometry of the reaction:

1 mole of aqueous solution of sodium sulfate produces 2 moles of sodium ions and 1 mole of sulfate ions.

Hence, the major ionic species present in the solution are sodium and sulfate ions.

You might be interested in

What pattern is caused by flammable liquids on the floor

NikAS [45]

The inverted cone, hope it helps .

6 0

3 years ago

What is the pOH of a 2.6 x 10-6 M H+ solution?

melomori [17]

Answer:

Approximately $8.41$ (assuming that the solution is at $\rm 25^\circ C$ , under which $K_{\rm w} = 10^{-14}$ .)

Explanation:

Let ${\rm [H^{+}]}$ and ${\rm [OH^{-}]}$ denote the concentration of $\rm H^{+}$ and $\rm OH^{-}$ respectively.

Let $K_{\rm w}$ denote the self-ionization constant of water. The exact value of $K_{\rm w}\!$ depends on the temperature of the solution. $K_{\rm w} =10^{-14}$ at $\rm 25^\circ C$ .

The product of ${\rm [H^{+}]}$ and ${\rm [OH^{-}]}$ in a solution (with $\rm M$ , or moles per liter, as the unit) is supposed to be equal to the $K_{\rm w}$ value of that solution at the corresponding temperature. In other words:

${\rm [H^{+}]} \cdot {\rm [OH^{-}]} = K_{\rm w}$ .

Rearrange to obtain an expression for ${[\rm OH^{-}]}$ :

$\begin{aligned}{\rm [OH^{-}]} &= \frac{K_{\rm w}}{[\rm H^{+}]}\end{aligned}$ .

Assume that the solution in this question is at $\rm 25^\circ C$ (for which $K_{\rm w} =10^{-14}$ .) For this solution:

$\begin{aligned}{\rm [OH^{-}]} &= \frac{K_{\rm w}}{[\rm H^{+}]} \\ &= \frac{10^{-14}}{2.6 \times 10^{-6}}\approx 3.85\times 10^{-9}\; \rm M\end{aligned}$ .

Hence, the $\rm pOH$ of this solution would be:

$\begin{aligned}\rm pOH &= -\log_{10}{\rm [OH^{-}]} \\&\approx -\log_{10} (3.85 \times 10^{-9}) \approx 8.41 \end{aligned}$ .

3 0

3 years ago

Who can write an essay

vampirchik [111]

Answer:

anyone you just have to know how

Explanation:

6 0

3 years ago

Materials with_____ bonds tend to be more easily scratched

iren [92.7K]

Materials with ionic bonds tend to be more easily scratched.

4 0

4 years ago

hich of the followings decrease the affinity of adult Hb with O2? A. 2,3 BPG B. hydrogen ion C. CO2 D. Ferrous ion

Ivahew [28]

Answer:

A. 2,3 BPG

Explanation:

2,3-bisphosphoglycerate (BPG), otherwise known as 2,3-DPG, enables the transition of hemoglobin from a very high-oxygen-affinity state to a reduced-oxygen-affinity state.

Tissues hemoglobin oxygen affinity is reduced by numerous physiological factors including.

1. Temperature Increased,

2. Carbon dioxide,

3. Acid and

4. 2,3-Bisphosphoglycerate (2,3-BPG)

all of which can contribute to decrease the oxygen affinity of hemoglobin which favours unloading and increased oxygen availability to our body cells.

3 0

3 years ago