Question 51. The degree of ionisation in water will be ……………. (a) 8 x 10 -7 (b) 0.8 x 10 -9 (c) 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 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.