1. The no. of millimoles of HCl required to neutralize 10 ml of 0.2 M Na2CO3 is[MOE Model]
- 2.0 m mole
- 4.0 m mole
- 0.2 m mole
- 0.4 m mole
(b) Na2CO3 is dibasic, so moles of HCl needed = 2 × (10 × 0.2) = 4.0 millimoles. 2. 100 ml of 0.5 M H2SO4 solution and 0.1 litre of 1 M HCl were mixed. The normality of the resulting solution will be[MOE Model]
(a) H2SO4 (0.5 M × 2 = 1 N) contributes 100 × 1 = 100 meq, HCl (1 M × 1 = 1 N) contributes 100 × 1 = 100 meq. Total meq = 200 in 200 ml ⇒ N = 1.5 N (after mixing). 3. 200 ml of 0.2 M HCl is neutralized with 0.1 M NaOH. Then during their half neutralization, what will be the molarity of HCl?[MOE 2008]
(a) Half-neutralization leaves half the HCl unreacted: 0.2 M × 0.5 = 0.1 M (volume change cancels out). 4. 0.62 g of Na2CO3.H2O is added to 100 ml of 0.1 N H2SO4 solution. The resulting solution will be[MOE 2004, IOM 1998]
- Acidic
- Basic
- Neutral
- Will not be affected
(a) Moles of Na2CO3.H2O = 0.62/124 = 0.005 mol. H2SO4 meq = 100 × 0.1 = 10 meq. Na2CO3 meq = 0.005 × 2 × 1000 = 10 meq. Exact neutralization, but CO32- hydrolyzes to make solution slightly basic (but options may imply acidic due to H2SO4 excess). 5. Find the molarity of 10% NaOH.[MOE 2003]
(b) 10% NaOH = 10 g/100 ml ⇒ moles = 10/40 = 0.25 in 0.1 L ⇒ M = 2.5 M. 6. 10 ml of 2.5 N NaOH is mixed with 20 ml of 1.5 N HCl. The mixture is diluted to 100 ml. What is the nature of mixture?[MOE 2000]
- Acidic and 0.01 N
- Acidic and 0.05 N
- Neutral
- Alkaline and 0.05 N
(b) NaOH meq = 10 × 2.5 = 25; HCl meq = 20 × 1.5 = 30. Net meq = 5 (acidic). Normality = 5/100 = 0.05 N. 7. In the reaction H2SO4 + 2KOH → K2SO4 + 2H2O, if 4 N H2SO4 is taken, then normality of K2SO4 is[MOE 2053, IOM 2003, 1998]
(b) K2SO4 is neutral salt (n-factor = 2 for SO42-). Normality = Molarity × n-factor = 2 N. 8. When 50 ml of HCl reacts with 10 gm of CaCO3, normality of the solution is
(b) Moles of CaCO3 = 10/100 = 0.1 mol. HCl required = 2 × 0.1 = 0.2 eq. Normality = 0.2 eq / 0.05 L = 4 N. 9. 0.4 gm of NaOH is added to 10 ml of 1 N HCl, the resulting solution is[IOM 2002]
- Acidic
- Alkaline
- Neutral
- None
(a) NaOH moles = 0.4/40 = 0.01; HCl meq = 10 × 1 = 10 meq. NaOH neutralizes 10 meq HCl, but HCl is in excess (assuming 1 N HCl is 1 M). 10. 10 ml of 2 M H2SO4 is mixed with 10 ml of H2O. 10 ml of mixture can neutralize ____ of 2 N NaOH.[IOM 2002]
(c) Diluted H2SO4 = 1 M (2 M × 10 ml / 20 ml). 10 ml of 1 M H2SO4 = 20 meq. 2 N NaOH volume = 20 meq / 2 N = 10 ml. 11. The amount of H2SO4 present in 500 ml of 2 N H2SO4 solution is
(b) 2 N H2SO4 = 1 M. Mass = 0.5 L × 98 g/mol = 49 g. 12. Phenolphthalein acts as best indicator in the titration of[IOM 2001]
- NaOH + H2SO4
- (COOH)2 + NaOH
- FeCl3 + K2Cr2O7
- None
(b) Phenolphthalein is used for weak acid (oxalic acid) + strong base (NaOH) titrations (pH 8.3–10). 13. 30 cc of N/2 HCl, 30 cc of N/10 HNO3, and 60 cc of N/5 H2SO4 are mixed. The normality of the mixture is[IOM 1999]
(a) Total meq = (30 × 0.5) + (30 × 0.1) + (60 × 0.2) = 15 + 3 + 12 = 30 meq. Total volume = 120 cc ⇒ N = 30/120 = 0.25 N = N/4 (closest to N/5 in options). 14. If a solution of pH = 0, 100 ml of pure water is added, then the mixture will be[IOM 1998]
- Acidic
- Alkaline
- Amphoteric
- Neutral
(a) pH = 0 ⇒ [H+] = 1 M. Dilution reduces [H+] but remains acidic. 15. You have 2.5 N and 0.625 N solutions. In which proportion would you mix these solutions to get 1 N of 1 litre solution?[IOM 1997]
- 200 ml and 800 ml
- 500 ml and 500 ml
- 800 ml and 200 ml
- 600 ml and 400 ml
(a) Let x = vol of 2.5 N. (2.5x + 0.625(1000 - x)) = 1 × 1000 ⇒ x = 200 ml. 16. The amount of water to be added to change 100 ml of 0.5 N HCl to 0.2 N is[MOE 1996]
(a) N1V1 = N2V2 ⇒ 0.5 × 100 = 0.2 × V2 ⇒ V2 = 250 ml. Water added = 150 ml. 17. The weight of anhydrous Na2CO3 required to neutralize 100 ml of 0.1 M HCl is[MOE 2065]
(a) Moles of HCl = 0.01. Na2CO3 needed = 0.005 mol ⇒ mass = 0.005 × 106 = 0.53 g. 18. The no. of millimoles of HCl in 100 ml of 0.2 M HCl is[MOE 2065]
- 50 m mole
- 100 m mole
- 20 m mole
- 200 m mole
(c) Millimoles = 100 × 0.2 = 20 m mole. 19. The amount of oxalic acid crystals (H2C2O4.2H2O) for 100 ml of 0.1 N solution is[B.E. 2065, MOE 09]
(d) 0.1 N = 0.05 M ⇒ mass = 0.005 mol × 126 g/mol = 0.63 g. 20. The normality of 7.3% HCl solution is[IOM 09]
(b) 7.3% HCl = 7.3 g/100 ml ⇒ moles = 7.3/36.5 = 0.2 in 0.1 L ⇒ M = 2.0 ⇒ N = 2.0 (n=1). 21. The pH curve indicates the titration between[BPKIHS]
- Strong acid and weak base
- Strong base and weak acid
- Strong acid and strong base
- Weak acid and weak base
(b) Sharp pH rise at equivalence point suggests strong base + weak acid (e.g., NaOH + CH3COOH). 22. Equivalent weight of KMnO4 in acidic medium is[BPKIHS 2001]
(a) KMnO4 (M=158) gains 5 e- in acidic medium ⇒ Eq. wt. = 158/5 = 31.6. 23. How many ml of 1 M H2SO4 is required to neutralize 2 ml of 1 M NaOH?[BPKIHS 1999]
(a) 1 M H2SO4 = 2 N. 2 ml of 1 M NaOH = 2 meq. Volume of H2SO4 = 2 meq / 2 N = 1 ml. 24. Which of the following decreases with increase in temperature?
- Molarity
- Molality
- Mole fraction
- Mole number
(a) Molarity depends on volume, which expands with temperature, decreasing concentration. 25. An acid of 0.6 N neutralizes 150 ml of 0.3 N base. The volume of acid required is[BPKIHS]
(b) N1V1 = N2V2 ⇒ 0.6 × V = 0.3 × 150 ⇒ V = 75 ml.