When Sodium( Na ) reacts with Water( H20 ) it gives Sodium hydroxide and hydrogen . So , first let me tell you something about water . Water consists of two radicals or ions - H+ and OH- . Since sodium is positively charged , it takes up the OH- ion and forms Sodium hydroxide( NaOH ) .
When Sodium( Na ) reacts with Water( H20 ) it gives Sodium hydroxide and hydrogen . So , first let me tell you something about water . Water consists of two radicals or ions - H+ and OH- . Since sodium is positively charged , it takes up the OH- ion and forms Sodium hydroxide( NaOH ) .
Lithium hydroxide is a strong base (alkali) and not a reducing agent, so it cannot reduce iodine. A reducing agent has to lose an electron to carry out reduction, e.g., Fe+2 - e = Fe+3. There is no scope for LiOH to lose an electron. Iodine generally is an oxidizing agent like oxidizing H2S to sulphur itself being reduced to hydroiodic acid. H2S + I2 =2HI + S. However, iodine can act as a reducing agent with a stronger oxidizing agent like chlorine forming ICl (iodine monochloride).
Lithium hydroxide is a strong base (alkali) and not a reducing agent, so it cannot reduce iodine. A reducing agent has to lose an electron to carry out reduction, e.g., Fe+2 - e = Fe+3. There is no scope for LiOH to lose an electron. Iodine generally is an oxidizing agent like oxidizing H2S to sulphur itself being reduced to hydroiodic acid. H2S + I2 =2HI + S. However, iodine can act as a reducing agent with a stronger oxidizing agent like chlorine forming ICl (iodine monochloride).
The reaction of iodine in base with phenylethanone (aka, acetophenone) is shown below. The general reaction is known as the iodoform reaction in which methyl ketones are converted to the corresponding carboxylic acid (here, benzoic acid) and triiodomethane (aka, iodoform). This is a classic qualitative analysis test as iodoform is a solid at room temp, not soluble in the medium and will precipitate out as a bright yellow colored material.
The reaction of iodine in base with phenylethanone (aka, acetophenone) is shown below. The general reaction is known as the iodoform reaction in which methyl ketones are converted to the corresponding carboxylic acid (here, benzoic acid) and triiodomethane (aka, iodoform). This is a classic qualitative analysis test as iodoform is a solid at room temp, not soluble in the medium and will precipitate out as a bright yellow colored material.
The reactions are same for [math]Cl_2[/math] and [math]Br_2[/math] .
I would like to add a few more things to the first (cold and dilute [math]NaOH[/math]) reaction. The products [math]NaI[/math] and [math]NaOI[/math] in that case can be visualized as the products of the reactions [math]HI + NaOH[/math] and [math]HOI + NaOH[/math] respectively. Non-metals tend to undergo disproportionation in basic mediums.
I would be impressed if you could now predict the products of the reaction between the reaction between [math]ICl + NaOH (hot and conc.)[/math]. I’m hiding the answer to this in the comment section.
The reactions are same for [math]Cl_2[/math] and [math]Br_2[/math] .
I would like to add a few more things to the first (cold and dilute [math]NaOH[/math]) reaction. The products [math]NaI[/math] and [math]NaOI[/math] in that case can be visualized as the products of the reactions [math]HI + NaOH[/math] and [math]HOI + NaOH[/math] respectively. Non-metals tend to undergo disproportionation in basic mediums.
I would be impressed if you could now predict the products of the reaction between the reaction between [math]ICl + NaOH (hot and conc.)[/math]. I’m hiding the answer to this in the comment section.
Yes. Ammonium ion can react with sodium hydroxide as ammonium hydroxide is a weaker base than sodium hydroxide. It reacts with sodium hydroxide to give ammonium hydroxide which furthur decomposes to evolve ammonia gas. This test is often used to detect the presence of ammonium ion. The reaction is as follows ——-
Yes. Ammonium ion can react with sodium hydroxide as ammonium hydroxide is a weaker base than sodium hydroxide. It reacts with sodium hydroxide to give ammonium hydroxide which furthur decomposes to evolve ammonia gas. This test is often used to detect the presence of ammonium ion. The reaction is as follows ——-
What I can tell is that NaOH is very soluble in water (1.09 Kg/L).
Apart from gas-gas mixtures that are miscible in all proportions, many other substances have a limit of solubility as shown in the example above. You cannot dissolve more than 1.09 kg per L of water at room temperature.
The nature of the solvent plays a big role in solubility: NaOH, being an ionic compound, will be soluble in a polar solvent but not in a covalent solvent.
The temperature may affect a little bit the solubility.
What I can tell is that NaOH is very soluble in water (1.09 Kg/L).
Apart from gas-gas mixtures that are miscible in all proportions, many other substances have a limit of solubility as shown in the example above. You cannot dissolve more than 1.09 kg per L of water at room temperature.
The nature of the solvent plays a big role in solubility: NaOH, being an ionic compound, will be soluble in a polar solvent but not in a covalent solvent.
The temperature may affect a little bit the solubility.
2 NaOH + NiCl2 ---> Ni(OH)2 (s) + 2 NaCl (aq) is one possible reaction, however the Nickle (II) hydroxide makes a green solid suspended in a clear solution which might turn it a pale green.
Another substance that can form a green gel precipitate that makes a solution appear pale green is the chromium 3+ ion. It makes a green gel like precipitate. Also if ammonia is added, then you can get it to dissolve possibly.
2 NaOH + NiCl2 ---> Ni(OH)2 (s) + 2 NaCl (aq) is one possible reaction, however the Nickle (II) hydroxide makes a green solid suspended in a clear solution which might turn it a pale green.
Another substance that can form a green gel precipitate that makes a solution appear pale green is the chromium 3+ ion. It makes a green gel like precipitate. Also if ammonia is added, then you can get it to dissolve possibly.
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When Sodium( Na ) reacts with Water( H20 ) it gives Sodium hydroxide and hydrogen . So , first let me tell you something about water . Water consists of two radicals or ions - H+ and OH- . Since sodium is positively charged , it takes up the OH- ion and forms Sodium hydroxide( NaOH ) .
Thanks for reading .
When Sodium( Na ) reacts with Water( H20 ) it gives Sodium hydroxide and hydrogen . So , first let me tell you something about water . Water consists of two radicals or ions - H+ and OH- . Since sodium is positively charged , it takes up the OH- ion and forms Sodium hydroxide( NaOH ) .
Thanks for reading .
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Lithium hydroxide is a strong base (alkali) and not a reducing agent, so it cannot reduce iodine. A reducing agent has to lose an electron to carry out reduction, e.g., Fe+2 - e = Fe+3. There is no scope for LiOH to lose an electron. Iodine generally is an oxidizing agent like oxidizing H2S to sulphur itself being reduced to hydroiodic acid. H2S + I2 =2HI + S. However, iodine can act as a reducing agent with a stronger oxidizing agent like chlorine forming ICl (iodine monochloride).
Lithium hydroxide is a strong base (alkali) and not a reducing agent, so it cannot reduce iodine. A reducing agent has to lose an electron to carry out reduction, e.g., Fe+2 - e = Fe+3. There is no scope for LiOH to lose an electron. Iodine generally is an oxidizing agent like oxidizing H2S to sulphur itself being reduced to hydroiodic acid. H2S + I2 =2HI + S. However, iodine can act as a reducing agent with a stronger oxidizing agent like chlorine forming ICl (iodine monochloride).
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The reaction of iodine in base with phenylethanone (aka, acetophenone) is shown below. The general reaction is known as the iodoform reaction in which methyl ketones are converted to the corresponding carboxylic acid (here, benzoic acid) and triiodomethane (aka, iodoform). This is a classic qualitative analysis test as iodoform is a solid at room temp, not soluble in the medium and will precipitate out as a bright yellow colored material.
The reaction of iodine in base with phenylethanone (aka, acetophenone) is shown below. The general reaction is known as the iodoform reaction in which methyl ketones are converted to the corresponding carboxylic acid (here, benzoic acid) and triiodomethane (aka, iodoform). This is a classic qualitative analysis test as iodoform is a solid at room temp, not soluble in the medium and will precipitate out as a bright yellow colored material.
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Here ya go:
3I2 + 6NaOH → 5NaI + NaIO3 + 3H2O
So, the products are sodium iodide (NaI), sodium iodate (NaIO3) and water (H2O).
Here ya go:
3I2 + 6NaOH → 5NaI + NaIO3 + 3H2O
So, the products are sodium iodide (NaI), sodium iodate (NaIO3) and water (H2O).
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It depends on whether the [math]NaOH[/math] taken is “cold and dilute” or “hot and concentrated”. The reactions involved are as follows:
[math]I_2 + 2NaOH (cold , dil.) ightarrow NaI + NaOI + H_{2}O[/math]
[math]3I_2 + 6NaOH (hot , conc.) ightarrow 5NaI + NaIO_3 + 3H_{2}O[/math]
The reactions are same for [math]Cl_2[/math] and [math]Br_2[/math] .
I would like to add a few more things to the first (cold and dilute [math]NaOH[/math]) reaction. The products [math]NaI[/math] and [math]NaOI[/math] in that case can be visualized as the products of the reactions [math]HI + NaOH[/math] and [math]HOI + NaOH[/math] respectively. Non-metals tend to undergo disproportionation in basic mediums.
I would be impressed if you could now predict the products of the reaction between the reaction between [math]ICl + NaOH (hot and conc.)[/math]. I’m hiding the answer to this in the comment section.
It depends on whether the [math]NaOH[/math] taken is “cold and dilute” or “hot and concentrated”. The reactions involved are as follows:
[math]I_2 + 2NaOH (cold , dil.)ightarrow NaI + NaOI + H_{2}O[/math]
[math]3I_2 + 6NaOH (hot , conc.) ightarrow 5NaI + NaIO_3 + 3H_{2}O[/math]
The reactions are same for [math]Cl_2[/math] and [math]Br_2[/math] .
I would like to add a few more things to the first (cold and dilute [math]NaOH[/math]) reaction. The products [math]NaI[/math] and [math]NaOI[/math] in that case can be visualized as the products of the reactions [math]HI + NaOH[/math] and [math]HOI + NaOH[/math] respectively. Non-metals tend to undergo disproportionation in basic mediums.
I would be impressed if you could now predict the products of the reaction between the reaction between [math]ICl + NaOH (hot and conc.)[/math]. I’m hiding the answer to this in the comment section.
More
VOTE
Yes. Ammonium ion can react with sodium hydroxide as ammonium hydroxide is a weaker base than sodium hydroxide. It reacts with sodium hydroxide to give ammonium hydroxide which furthur decomposes to evolve ammonia gas. This test is often used to detect the presence of ammonium ion. The reaction is as follows ——-
Yes. Ammonium ion can react with sodium hydroxide as ammonium hydroxide is a weaker base than sodium hydroxide. It reacts with sodium hydroxide to give ammonium hydroxide which furthur decomposes to evolve ammonia gas. This test is often used to detect the presence of ammonium ion. The reaction is as follows ——-
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This question is not clear?
What I can tell is that NaOH is very soluble in water (1.09 Kg/L).
Apart from gas-gas mixtures that are miscible in all proportions, many other substances have a limit of solubility as shown in the example above. You cannot dissolve more than 1.09 kg per L of water at room temperature.
The nature of the solvent plays a big role in solubility: NaOH, being an ionic compound, will be soluble in a polar solvent but not in a covalent solvent.
The temperature may affect a little bit the solubility.
This question is not clear?
What I can tell is that NaOH is very soluble in water (1.09 Kg/L).
Apart from gas-gas mixtures that are miscible in all proportions, many other substances have a limit of solubility as shown in the example above. You cannot dissolve more than 1.09 kg per L of water at room temperature.
The nature of the solvent plays a big role in solubility: NaOH, being an ionic compound, will be soluble in a polar solvent but not in a covalent solvent.
The temperature may affect a little bit the solubility.
More
VOTE
2 NaOH + NiCl2 ---> Ni(OH)2 (s) + 2 NaCl (aq) is one possible reaction, however the Nickle (II) hydroxide makes a green solid suspended in a clear solution which might turn it a pale green.
Another substance that can form a green gel precipitate that makes a solution appear pale green is the chromium 3+ ion. It makes a green gel like precipitate. Also if ammonia is added, then you can get it to dissolve possibly.
3 NaOH + CrCl3 ---> Cr(OH)3 (s) + 3 NaCl (aq)
2 NaOH + NiCl2 ---> Ni(OH)2 (s) + 2 NaCl (aq) is one possible reaction, however the Nickle (II) hydroxide makes a green solid suspended in a clear solution which might turn it a pale green.
Another substance that can form a green gel precipitate that makes a solution appear pale green is the chromium 3+ ion. It makes a green gel like precipitate. Also if ammonia is added, then you can get it to dissolve possibly.
3 NaOH + CrCl3 ---> Cr(OH)3 (s) + 3 NaCl (aq)
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