{"id":16154,"date":"2023-12-10T04:48:59","date_gmt":"2023-12-09T23:18:59","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=16154"},"modified":"2023-12-11T17:14:52","modified_gmt":"2023-12-11T11:44:52","slug":"selina-icse-solutions-class-10-chemistry-analytical-chemistry-uses-ammonium-hydroxide-sodium-hydroxide","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/selina-icse-solutions-class-10-chemistry-analytical-chemistry-uses-ammonium-hydroxide-sodium-hydroxide\/","title":{"rendered":"Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide"},"content":{"rendered":"

Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide<\/span><\/h2>\n

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Chemistry Chapter 4 Analytical Chemistry: Uses Of Ammonium Hydroxide And Sodium Hydroxide. You can download the Selina Concise Chemistry ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Chemistry for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.<\/p>\n

Download Formulae Handbook For ICSE Class 9 and 10<\/a><\/p>\n

ICSE Solutions<\/a>Selina ICSE Solutions<\/a><\/p>\n

Selina ICSE Solutions for Class 10 Chemistry Chapter 4 Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide<\/strong><\/p>\n

Exercise 1<\/strong><\/span><\/p>\n

Solution 1.<\/strong><\/span><\/p>\n

(a) Ferrous salts : Light green
\n(b) Ammonium salts : Colourless
\n(c) Cupric salts : Blue
\n(d) Calcium salts : Colourless
\n(e) Aluminium salts : Colourless<\/p>\n

Solution 2.<\/strong><\/span><\/p>\n

(a) Cu(OH)2
\n<\/sub>(b)\u00a0ZnO
\n(c)\u00a0NaOH
\n(d) NH4<\/sub>OH
\n(e) Na+<\/sup>, Ca2+
\n<\/sup>(f) Fe2+<\/sup>, Mn2+
\n<\/sup>(g)\u00a0Aluminium
\n(h) Zn(OH)2<\/sub>\u00a0and Al(OH)3
\n<\/sub>(i)\u00a0PbO
\n(j) Ammonium ion<\/p>\n

Solution 3.<\/strong><\/span><\/p>\n

\"Selina<\/p>\n

Solution 4.<\/strong><\/span><\/p>\n

\"Selina<\/p>\n

Solution 5.<\/strong><\/span><\/p>\n

\"Selina<\/p>\n

Solution 6.<\/strong><\/span><\/p>\n

\"Selina<\/p>\n

Solution 7.<\/strong><\/span><\/p>\n

(a) ZnCl2
\n<\/sub>(b) Zn(OH)2<\/sub><\/p>\n

Solution 8.<\/strong><\/span><\/p>\n

(a)\u00a0PbO
\n(b)\u00a0ZnO
\n(c) K2<\/sub>ZnO2<\/sub><\/p>\n

Solution 9.<\/strong><\/span><\/p>\n

\"Selina<\/p>\n

Solution 10.<\/strong><\/span><\/p>\n

When freshly precipitated aluminum hydroxide reacts with caustic soda solution,\u00a0whitesalt<\/span>\u00a0of sodium meta\u00a0aluminate<\/span>\u00a0is obtained.
\n\"Selina<\/p>\n

Solution 11.<\/strong><\/span>
\n\"Selina<\/p>\n

Solution 12.<\/strong><\/span><\/p>\n

\"Selina
\nWith excess of\u00a0NaOH<\/span>, white gelatinous ppt. of Zn (OH)2<\/sub><\/span>\u00a0is soluble. So, these two cannot be distinguished by\u00a0NaOH<\/span>\u00a0alone. However white ppt. of\u00a0Pb<\/span><\/span>(<\/span>OH)2\u00a0<\/sub>is readily soluble in acetic acid also.
\n\"Selina
\n(i) On addition of NH4<\/sub>OH to calcium salts no precipitation of\u00a0Ca(OH)2\u00a0<\/sub>occurs even with addition of excess of NH4<\/sub>OH because the concentration of OH–<\/sup>ions from ionization of NH4<\/sub>OH is so low that it cannot precipitate the hydroxide of calcium.
\nPb(NO3<\/sub>)2<\/sub>+2 NH4<\/sub>OHPb(OH)2<\/sub>+2NH4<\/sub>NO3<\/sub>
\n\"Selina
\nSolution 13.<\/strong><\/span><\/p>\n

Lead carbonate is dissolved in dilute nitric acid and then ammonium hydroxide is added to it. A white precipitate is formed which is insoluble in excess.
\nZinc carbonate is dissolved in dilute nitric acid and then ammonium hydroxide is added to it. A white precipitate is formed which is soluble in excess.<\/p>\n

Solution 14.<\/strong><\/span><\/p>\n

Reagent bottles A and B can identified by using calcium salts such as\u00a0Ca(NO3<\/sub>)2<\/sub>.<\/p>\n

On adding\u00a0NaOH\u00a0to Ca (NO3<\/sub>)2<\/sub>, Ca (OH)2<\/sub>\u00a0is precipitated as white precipitate which is sparingly soluble in excess of\u00a0NaOH.
\nCa(NO3<\/sub>)2<\/sub>+2NaOH \u2192 Ca(OH)2<\/sub>+ 2NaNO3
\n<\/sub><\/p>\n

Whereas, on addition of NH4<\/sub>OH to calcium salts, no precipitation of Ca(OH)2\u00a0<\/sub>occurs even with addition of excess of NH4<\/sub>OH because the concentration of OH–<\/sup>ions from the ionization of NH4<\/sub>OH is so low that it cannot precipitate the hydroxide of calcium.
\nSo the reagent bottle which gives white precipitate is\u00a0NaOH\u00a0and the other is NH4<\/sub>OH.<\/p>\n

Intext Exercise<\/strong><\/span><\/p>\n

Solution 1.<\/strong><\/span><\/p>\n

(i) Analysis: The determination of chemical components in a given sample is called analysis.
\n(ii) Qualitative analysis: The analysis which involves the identification of the unknown substances in a given sample is called qualitative analysis.
\n(iii) Reagent: A reagent is a substance that reacts with another substance.
\n(iv) Precipitation: It is the process of formation of an insoluble solid when solutions are mixed. The solid thus formed is called precipitate.<\/p>\n

Solution 2.<\/strong><\/span><\/p>\n

(i) Yellow
\n(ii) Colourless
\n(iii) PaleGreen
\n(iv) Colourless
\n(v) Colourless<\/p>\n

Solution 3.<\/strong><\/span><\/p>\n

(i) Fe3+
\n<\/sup>(ii) Cu2+
\n<\/sup>(iii) Cu+2
\n<\/sup>(iv) Mn2+<\/sup><\/p>\n

Solution 4.<\/strong><\/span><\/p>\n

(i) Ca(OH)2
\n<\/sub>(ii) Fe(OH)2<\/sub>\u00a0and Cu(OH)2
\n<\/sub>(iii) Zn(OH)2<\/sub>\u00a0and\u00a0Pb(OH)2<\/sub><\/p>\n

Solution 5.<\/strong><\/span><\/p>\n

\"Selina<\/p>\n

Solution 6.<\/strong><\/span><\/p>\n

NH4<\/sub>OH and\u00a0NaOH\u00a0can be distinguished by using calcium salts.<\/p>\n

For example on adding\u00a0NaOH\u00a0to\u00a0Ca(NO3<\/sub>)2<\/sub>, Ca(OH)2<\/sub>\u00a0is obtained as white precipitate which is sparingly soluble in excess of\u00a0NaOH.<\/p>\n

Ca(NO3<\/sub>)2<\/sub>\u00a0+\u00a02NaOH\u00a0\u2192 Ca(OH)2<\/sub>\u00a0+\u00a02NaNO3<\/sub><\/p>\n

On addition of NH4<\/sub>OH to calcium salts, no precipitation of\u00a0Ca(OH)2<\/sub>\u00a0occurs even with the addition of excess of NH4<\/sub>OH.This is because the concentration of OH–<\/sup>\u00a0ions from the ionization of NH4<\/sub>OH is so low that it cannot precipitate the hydroxide of calcium.<\/p>\n

Solution 7.<\/strong><\/span><\/p>\n

(i) Fe(OH)2\u00a0<\/sub>and\u00a0Pb(OH)2
\n<\/sub>(ii) Cu(OH)2\u00a0<\/sub>and\u00a0<\/sub>Zn(OH)2<\/sub><\/p>\n

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