Free Acids, Bases and Salts MCQs with Answers

241. Suppose Kb value for HSO4- is 1 x 10^-9 what is the Ka value for H2SO4?

• A. 1 x 10^-9
• B. 1 x 10^-5
• C. 1 x 10^-3
• D. 1 x 10^-7

Explanation: To find the Ka for H2SO4, we use the relationship Ka × Kb = Kw, where Kw is the ion product of water, equal to 1 × 10-14. Given that Kb for HSO4- is 1 × 10-9, we can rearrange the equation to find Ka: Ka = Kw / Kb = 1 × 10-14 / 1 × 10-9 = 1 × 10-5. Therefore, the correct answer is Option B. Other options are incorrect because they do not satisfy this calculation.

242. The strength of a mixture of HCl & H2SO4 is 0.1 N. On treatment with an excess of AgNO3 solution, 20 ml of this acid mixture gives 0.1435 gm of AgCl. The strength of the H2SO4 is:

• A. 24.5 g/litre
• B. 2.45 g/litre
• C. 4.9 g/litre
• D. 49 g/litre

Explanation: To find the strength of H2SO4, first calculate the moles of AgCl formed: the molar mass of AgCl is approximately 143.35 g/mol, so 0.1435 g corresponds to 0.001 moles of AgCl. Since AgCl is formed from Cl- ions from HCl, calculate the moles of HCl in the solution: 0.001 moles of Cl- correspond to 0.001 moles of HCl. Given that the total normality of the acid mixture is 0.1 N, and assuming all of this is due to H+ ions from HCl and H2SO4, we can set up an equation based on the stoichiometry of the reactions:2HCl + 2AgNO3 → 2AgCl + 2HNO3H2SO4 + 2AgNO3 → Ag2SO4 + 2HNO3Given the 0.001 moles of HCl, the rest of the 0.1 N is contributed by H2SO4. Therefore, calculate the concentration of H2SO4 using the formula: Concentration (g/litre) = Normality × Equivalent weight = 0.049 x 49 = 2.45 g/litre. Thus, the correct answer is 2.45 g/litre.

243. If x gm is the mass of NaHC2O4 required to neutralize 100 ml of 0.2 (M) NaOH and y gm that is required to reduce 100 ml of 0.02 (M) KMnO4 in acidic medium, then:

• A. x = y
• B. 2x = y
• C. x = 4y
• D. 4x = y

Explanation: To solve this problem, we need to consider the stoichiometry of the reactions involved:For neutralizing NaOH, the reaction is: NaHC2O4 + NaOH → Na2C2O4 + H2O. Here, 1 mole of NaHC2O4 neutralizes 1 mole of NaOH.For reducing KMnO4 in acidic medium, the reaction is: 5 NaHC2O4 + 2 KMnO4 + 3 H2SO4 → 5 CO2 + 2 MnSO4 + K2SO4 + 8 H2O. Here, 5 moles of NaHC2O4 reduce 2 moles of KMnO4.From the stoichiometry, it is clear that the mass of NaHC2O4 needed to neutralize NaOH (x) is four times the mass needed to reduce KMnO4 (y), hence x = 4y. Other options do not align with the stoichiometric calculations of these reactions.

244. 100 ml of each of 0.5 N NaOH, N/5 HCl and N/10 H2SO4 are mixed together. The resulting solution will be:

• A. Acidic
• B. Neutral
• C. Alkaline
• D. None

Explanation: To determine the nature of the resulting solution, calculate the total equivalents of H+ and OH- ions. The 0.5 N NaOH provides 0.05 equivalents of OH- ions in 100 ml. The N/5 HCl provides 0.02 equivalents of H+ ions, and N/10 H2SO4 contributes 0.01 equivalents of H+ ions, totaling 0.03 equivalents of H+ ions. Since 0.05 equivalents of OH- are greater than the 0.03 equivalents of H+, the solution is alkaline. Options A and B are incorrect as they do not account for the excess OH- ions. Option D is incorrect because the solution clearly shows an alkaline nature.

245. 100 ml of N/5 NaOH will neutralize:

• A. 0.0618g of H3BO3
• B. 0.1855g of H3BO3
• C. 1.2368g of H3BO3
• D. 0.03092g of H3BO3

Explanation: The correct answer is 1.2368g of H3BO3. The reaction between NaOH and H3BO3 is a neutralization reaction, where the base (NaOH) reacts with the acid (H3BO3) to form water and a salt. Using the normality of NaOH (N/5 or 0.2N) and the volume (100 ml or 0.1 L), we calculate the equivalent moles of NaOH, which is 0.02 equivalents. H3BO3 has a molar mass of 61.83 g/mol, and since it provides one equivalent per mole in the reaction, the equivalent weight is the same as its molar mass. Thus, 0.02 equivalents of H3BO3 corresponds to 0.02 x 61.83 = 1.2368 g, which can be neutralized by 100 ml of N/5 NaOH. The other options do not align with this stoichiometric requirement.

246. Which of the following is/are soft bases?H2O, H-, CO,CO2, C2H4, CN-

• A. C2H4
• B. H- ,CN- , CO
• C. H-, CN-, CO, C2H4
• D. H-, CO, CO2, C2H4

Explanation: Soft bases have the donor atom of highly polarisable and they preferably combine with the metal ions of lower oxidation states.

247. Which of the following constitutes a set amphiprotic species?

• A. H3O+, H2PO4- , HCO-3
• B. H2O , HPO4-2 , H2PO2-
• C. H2O, H2PO3- , HPO42-
• D. HC2O4-, H2PO4-, SO42-

Explanation: H3O+ cannot take up proton; H2PO2- cannot give up proton, SO2− 4 cannot give proton.

248. Which of the following order represent the order for the strength of base?

• A. CH3CH2-> NH− 2 > HC ≡ C- > OH-
• B. H - C ≡ C- > CH3CH-2 > NH−2 > OH-
• C. OH- > NH-2 > HC ≡ C- > CH3CH2-
• D. NH2-> HC ≡ C- > OH- > CH3CH2-

Explanation: The strength of acid is in the order of CH3 - CH3 < NH3 < HC ≡ CH < H2O

249. The pH of 0.1 (M) solution of the following salts increasesin the order:

• A. NaCl < NH4Cl < NaCN < HCl
• B. HCl < NH4Cl < NaCl < NaCN
• C. NaCN < NH4Cl < NaCl < HCl
• D. HCl < NaCl < NaCN < NH4Cl

Explanation: pH of NaCl = 7 ; pH of NH4Cl < 7 ; pH of NaCN > 7 and pH of 0.1 (M) HCl = 1

250. A solution of ammonium cyanide is :

• A. acidic in nature
• B. alkaline in nature
• C. neutral in nature
• D. amphoteric in nature

Explanation: A solution of NH4CN is alkaline because CN- is more strong conjugate base than NH4 + , conjugate acid. The hydrolysis of CN- proceeds more that of NH4+.