You to definitely litre away from drinking water consists of ten mol hydrogen ions

You to definitely litre away from drinking water consists of ten 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 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

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