# That litre off liquids includes 10 mol hydrogen ions

28.06.2022 No Comments

That litre off liquids 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 three.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. 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 the pH of 1 M CH3. Ka of acetic escort Sacramento CA 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

= 2.35 x ten -5 pOH = cuatro.62 pH + pOH = fourteen pH = fourteen – cuatro.62 = 9.38

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.

Concern 61. This new pH out-of 0.001 Meters NaOH is …………. (a) 3 (b) 2 (c) 11 (d) several Respond to: (c) 11 Solution: 0.001 M NaOH setting [OH – ] 0.001 . ten -3 pOH = step three pH + pOH = 14 pH = 14 – step 3 = 11