Chemistry, 08.12.2019 22:31 jjsavage06

How many ions are present in 30.0ml of 0.600m na2co3 solution?

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Answer from: Skye251

$\boxed{{\text{3}}{\text{.25}}\times{\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ ions}}}$ are present in 30 mL of 0.600 M ${\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}$ solution.

Further Explanation:

The proportion of substance in the mixture is called concentration. The most commonly used concentration terms are as follows:

1. Molarity (M)

2. Molality (m)

3. Mole fraction (X)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

Molarity is a concentration term that is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by M and its unit is mol/L.

The formula to calculate the molarity of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ ] solution is as follows:

${\text{Molarity of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{ solution}}=\frac{{{\text{Moles}}\;{\text{of}}\;{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}}{{{\text{Volume }}\left({\text{L}}\right){\text{ of}}\;{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{ solution}}}}$ …… (1)

Rearrange equation (1) to calculate the moles of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ solution.

$\begin{aligned}{\text{Moles}}\;{\text{of}}\;{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}=\left({{\text{Molarity of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{ solution}}}\right)\\\left({{\text{Volume of}}\;{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{ solution}}}\right)\\\end{aligned}$ …… (2)

The molarity of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ solution is 0.600 M.

The volume of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ solution is 30 mL.

Substitute these values in equation (2).

$\begin{aligned}{\text{Moles}}\;{\text{of}}\;{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}&=\left({{\text{0}}{\text{.600 M}}}\right)\left({{\text{30 mL}}}\right)\left({\frac{{{{10}^{ - 3}}{\text{ L}}}}{{{\text{1 mL}}}}}\right)\\&=0.01{\text{8 mol}}\\\end{aligned}$

The dissociation of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ occurs as follows:

${\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}\rightleftharpoons 2{\text{N}}{{\text{a}}^ + } + {\text{CO}}_3^{2 - }$

This indicates that one mole of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ dissociates to form two moles of ${\text{N}}{{\text{a}}^ + }$ ions and one mole of ${\text{CO}}_3^{2 - }$ ion. So total of three moles of ions are produced by one mole of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ .

The total number of moles of ions produced during the reaction is calculated as follows:

$\begin{aligned}{\text{Moles of ions}}&=\left({{\text{0}}{\text{.018 mol}}}\right)\left({\text{3}} \right)\\&=0.0{\text{54 mol}}\\\end{aligned}$

The formula to calculate the number of ions is as follows:

${\text{Number of ions}}=\left({{\text{Moles of ions}}}\right)\left({{\text{Avogadro's Number}}}\right)$ …… (3)

The number of moles of ions is 0.054 mol.

The value of Avogadro’s number is ${\text{6}}{\text{.022}}\times{\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{ions}}$ .

Substitute these values in equation (3).

$\begin{aligned}{\text{Number of ions}}{\mathbf{ &= }}\left({0.054{\text{ mol}}}\right)\left( {\frac{{{\text{6}}{\text{.022}}\times{\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ ions}}}}{{{\text{1 mol}}}}} \right)\\&={\text{3}}{\text{.25188}}\times{\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ ions}}\\&\approx {\text{3}}{\text{.25}}\times{\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ ions}}\\\end{aligned}$

Learn more:

1. The mass of ethylene glycol: link

2. Calculate the moles of ions in HCl solution: link

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration terms

Keywords: molarity, ions, moles, Na2CO3, Na+, CO32-, 2Na+, 0.018 mol, 0.054 mol, Avogadro’s number, 3.25*10^22 ions, volume of Na2CO3, moles of Na2CO3.

Answer from: Quest

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Answer from: Quest

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.