You to litre regarding water includes 10 mol hydrogen ions

Question 51. The degree of ionisation in water will be ……………. (a) 8 x 10 -7 (b) 0.8 x 10 -9 (c) step step 3.six x 10 -7 (d) 3.6 x 10 -9 Answer: (a) 8 x 10 -7 Solution: 1 litre of water contains mole. So, degree of ionisation = \(\frac <10^<-7>\times 18><1000>\) = 1.8 x 10 -7

COOH solution?

Question 52. If the solubility product of lead iodide (PbI₂) is 3.2 x 10 -8 . Then its solubility in moles/litre will be …………. (a) 2 x 10 -3 . (b) cuatro x 10 -4 (c) 1.6 x 10 -5 (d) 1.8 x 10 -5 Solution: K_sp = 4s 3 4s 3 = 3.2 x 10 -8 s = 2 x 10 -3 M

Matter 54

Question 53. The pH of a soft drink is 3.82. It’s hydrogen ion concentration will be …………… (a) 1.96 x 10 -2 mol / L (b) 1.96 x 1o -3 mol / L (c) 1.5 x 10 -4 mol / L (d) 1.96 x 10 -1 mol / L Answer: (c) 1.5 x 10 -4 mol / L Solution: pH = 3.82 = – log₁₀[H + ] ? [H + ] = 1.5 x 10 -4 mol / litre

The pH of a solution at 25°C containing 0.10 M sodium acetate and 0.03 M acetic acid is ………….. (pK_a for CH₃COOH = 4.57) (a) 4.09 (b) 5.09 (c) 6.10 (d) 7.09 Answer: (b) 5.09 Solution: pH = pK_a + log \(\frac < [salt]>< [acid]>\) = 4.57 + log \(\frac < 0.10>< 0.03>\) = 5.09

Question 55. A weak acid is 0.1% ionised in 0.1 M solution. Its pH is ………….. (a) 2 (b) 3 (c) 4 (d) 1 Answer: (c) 4 Solution: For a monobasic acid [H + ] = c.? = \(\frac < 1>< 2>\) x 0 https://datingranking.net/escort-directory/spokane/.001 = 10 -4 pH = – log₁₀[10 -4 ] = 4

Question 56. Which one of the following is not a buffer solution? (a) 0.8 M H₂S + 0.8 M KHS. (b) 2 M C₆H₅NH₂ + 2 M C₆H₅N (c) 3 M H₂CO₃ + 3 M KHCO₃ (d) 0.05 M KCIO₄ + 0.05 M HCIO Answer: (d) 0.05 M KCIO₄ + 0.05 M HCIO Hint. HClO₄ is a strong acid while buffer is a mixture of weak acid and its salt.

Question 57. The pH of pure water or neutral solution at 50°C is …………… (pK_w = at 50°C) (a) 7.0 (b) 7.13 (c) 6.0 (d) 6.63 Answer: (d) 6.63 Solution: [H + ] [OH – ] = 10 – [H + ] = [OH – ] [H + ] = \(\frac < <>^< \frac> < 2>> >< 2>\) ? pH = 6.63

Question 59. What is the pH of 1 M CH₃. Ka of acetic acid is 1.8 x 10 -5 . K = 10 -14 mol 2 litre 2 . (a) 9.4 (b) 4.8 (c) 3.6 (d) 2.4 Answer: (a) 9.4 Solution: CH₃COO + H₂O \(\rightleftharpoons\) CH₃COOH + OH – [OH – ] = c x h

= 2.thirty five x 10 -5 pOH = 4.62 pH + pOH = 14 pH = fourteen – cuatro.62 = nine.38

Question 60. 4Na + O₂ > 2Na₂O Na₂O + H₂O > 2NaOH In the given reaction, the oxide of sodium is ………….. (a) Acidic (b) Basic (c) Amphoteric (d) Neutral Answer: (b) Basic Solution. Na₂O form NaOH so that it is basic oxide.

Question 61. The pH of 0.001 Yards NaOH would-be …………. (a) step 3 (b) 2 (c) 11 (d) twelve Address: (c) 11 Solution: 0.001 Yards NaOH means [OH – ] 0.001 . ten -3 pOH = step 3 pH + pOH = fourteen pH = fourteen – step three = 11