Back titration and hardness of water

35 questions on Back titration and hardness of water from the Equivalent Concept and Titration chapter: 5 NEET level, 12 JEE Main, 18 JEE Advanced. Each one opens with its answer and solution.

• Which of the following salts is mainly responsible for permanent hardness of water? NEET level
• Hardness is expressed as ppm of CaCO3 (molar mass 100 g mol^-1), where 1 ppm = 1 mg per litre. A water sample containing 1.0x10^-3 mol/L of Ca^2+ ions (and no o... NEET level
• 0.600 g of pure CaCO3 (molar mass 100 g mol^-1) is treated with 100 mL of 0.200 M HCl. The excess acid is back-titrated with 0.100 M NaOH. The volume of NaOH ne... NEET level
• A water sample contains 97.2 mg/L of Ca(HCO3)2 (molar mass 162) and 60 mg/L of MgSO4 (molar mass 120). Hardness is expressed as ppm of CaCO3 (molar mass 100). T... JEE Main level
• 50.0 mL of hard water requires 12.4 mL of 0.0100 M EDTA (1 : 1 complexation with Ca^2+ and Mg^2+). The total hardness in ppm of CaCO3 (molar mass 100) is JEE Main level
• A 0.900 g aspirin tablet (acetylsalicylic acid, C9H8O4, molar mass 180) is boiled with 50.0 mL of 0.500 M NaOH, which neutralises and hydrolyses it (2 mol NaOH ... JEE Main level
• A 0.800 g antacid tablet containing Mg(OH)2 (molar mass 58) is dissolved in 50.0 mL of 0.200 M HCl. The excess acid needs 20.0 mL of 0.150 M NaOH. Assuming no o... JEE Main level
• Which of the following statements about hardness of water are correct? JEE Main level
• CaCO3 is dissolved by adding a known excess of standard HCl, and the excess acid is titrated with standard NaOH. Which statements are correct? JEE Main level
• A water sample has total hardness due to divalent ions of 3.00x10^-3 mol/L. Taking CaCO3 = 100 and Na2CO3 = 106 (all hardness precipitated as carbonates, 1 : 1)... JEE Main level
• 50.0 mL of water containing only Na2CO3 and NaHCO3 as alkalinity is titrated with 0.0200 M HCl. The phenolphthalein end point (CO3^2- -> HCO3-) is reached at 3.... JEE Advanced level
• In a Kjeldahl analysis, ammonia from 1.000 g of a food sample is absorbed in 40.0 mL of 0.1000 M H2SO4. The excess acid (both protons) needs 22.1 mL of 0.2000 M... JEE Advanced level
• Water contains 48.6 mg/L of Ca(HCO3)2 (molar mass 162) and 72 mg/L of MgSO4 (molar mass 120). It is softened by pure lime, Ca(OH)2 (molar mass 74), which reacts... JEE Advanced level
• An EDTA solution is made by dissolving 0.930 g of disodium EDTA dihydrate (molar mass 372) in water to 250 mL. 50.0 mL of raw water needs 14.2 mL of this EDTA. ... JEE Advanced level
• 0.920 g of a mixture of CaCO3 (molar mass 100) and MgCO3 (molar mass 84) is heated with 50.0 mL of 0.500 M HCl. The excess acid needs 5.0 mL of 1.00 M NaOH. The... JEE Advanced level
• 50.0 mL of a water sample whose only anions are chloride and sulphate is passed through a cation-exchange resin in H+ form, which exchanges every Ca^2+ or Mg^2+... JEE Advanced level
• 25.0 mL of a solution containing NaOH and Na2CO3 is titrated with 0.100 M HCl. With phenolphthalein the end point is at 18.0 mL; adding methyl orange, a further... JEE Advanced level
• 0.400 g of pyrolusite ore is heated with 50.0 mL of 0.100 M Na2C2O4 in dilute acid (MnO2 + C2O4^2- + 4H+ -> Mn^2+ + 2CO2 + 2H2O). The unreacted oxalate needs 20... JEE Advanced level
• Temporary hardness of water is caused by the presence of: NEET level
• In the EDTA titration of hard water using ammonia buffer (pH about 10) and Eriochrome Black T, the colour change at the end point is: NEET level
• 0.800 g of impure CaCO3 is heated with 50.0 mL of 0.200 M HCl (excess). The unreacted acid needs 12.0 mL of 0.250 M NaOH. Taking CaCO3=100, the percentage of Ca... JEE Main level
• In Kjeldahl's method, the ammonia from 0.600 g of an organic compound is absorbed in 40.0 mL of 0.100 M H2SO4. The excess acid needs 18.0 mL of 0.200 M NaOH. Ta... JEE Main level
• 50.0 mL of a hard water sample is titrated with 0.0105 M EDTA (buffer pH 10, EBT) and requires 11.3 mL. Taking CaCO3=100 and EDTA to metal ion as 1:1, find the ... JEE Main level
• A tablet sample of mass 0.50 g containing aspirin (acetylsalicylic acid, C9H8O4, molar mass 180) is boiled with 30.0 mL of 0.200 M NaOH so that each aspirin mol... JEE Main level
• Which statements about hardness of water are correct? JEE Main level
• One litre of water contains 40.5 mg of Ca(HCO3)2 (molar mass 162) and 73.0 mg of Mg(HCO3)2 (molar mass 146). Clark's process uses Ca(OH)2 (molar mass 74): Ca(HC... JEE Advanced level
• 0.600 g of pyrolusite ore is heated with 50.0 mL of 0.100 M oxalic acid and dilute H2SO4: MnO2 + H2C2O4 + 2H+ -> Mn^2+ + 2CO2 + 2H2O. The unreacted oxalic acid ... JEE Advanced level
• To estimate Al^3+, 25.0 mL of the sample is treated with 40.0 mL of 0.0500 M EDTA (excess). The unreacted EDTA is back-titrated with 0.0400 M Zn^2+, requiring 1... JEE Advanced level
• Two 100.0 mL portions of a hard water are analysed with 0.0100 M EDTA (EBT, pH 10). The first (raw water) needs 17.5 mL. The second is boiled, filtered and made... JEE Advanced level
• 1.500 g of a fertiliser containing (NH4)2SO4 (molar mass 132) is boiled with excess NaOH; the ammonia is absorbed in 50.0 mL of 0.400 M HCl. The unreacted HCl n... JEE Advanced level
• Each litre of a water sample contains 48.6 mg Ca(HCO3)2, 33.3 mg CaCl2 and 31.2 mg MgSO4. Washing soda treatment removes only the permanent hardness, each mole ... JEE Advanced level
• 0.900 g of powdered eggshell (assume only CaCO3 reacts) is treated with 50.0 mL of 0.250 M HCl and the mixture is diluted to exactly 250.0 mL. A 25.0 mL portion... JEE Advanced level
• 1.00 g of marble chips (impure CaCO3) is dissolved in 40.0 mL of 0.500 M HCl (excess). The unreacted acid is back-titrated with 0.250 M NaOH and needs 22.0 mL. ... JEE Advanced level
• A water sample contains 60.0 mg of Ca^2+ and 24.0 mg of Mg^2+ per litre (Ca=40, Mg=24, CaCO3=100). A 100.0 mL portion is titrated with 0.0100 M EDTA. At pH 10 b... JEE Advanced level
• 25.0 mL of a solution of Na2CO3 and NaHCO3 is titrated with 0.100 M HCl. The phenolphthalein end point (CO3^2- -> HCO3^-) is reached after 12.0 mL; a further 18... JEE Advanced level