Question 51. The degree of ionisation in water will be ……………. (a) 8 x 10 -7 (b) 0.8 x 10 -9 (c) step step three.6 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 provider?
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: Ksp = 4s 3 4s 3 = 3.2 x 10 -8 s = 2 x 10 -3 M
Question 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 ………….. (pKa for CH3COOH = 4.57) (a) 4.09 (b) 5.09 (c) 6.10 (d) 7.09 Answer: (b) 5.09 Solution: pH = pKa + 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 0.001 = 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 …………… (pKw = 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 escort service Killeen 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 ten -5 pOH = cuatro.62 pH + pOH = 14 pH = fourteen – 4.62 = 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.
Matter 61. New pH off 0.001 Meters NaOH would be …………. (a) step three (b) 2 (c) eleven (d) several Answer: (c) eleven Solution: 0.001 M NaOH setting [OH – ] 0.001 . 10 -3 pOH = step three pH + pOH = fourteen pH = fourteen – step three = eleven