Brand new pH off an example of vinegar is actually step step three

Question 13. 76. Calculate the concentration of hydrogen ion in it. Answer: pH = – log_ten [H₃O + ] = – log₁₀ = – pH = – 3.76 = \(\overline\).dos4 [H₃O + ] = antilog \(\overline<4>\).24 = l.738 x 10 -4 [H₃O + ] = 1.74.x 10 -4 M

Question 14. The ionisation constant of HF, HCOOH, HCN at 298 K are 6.8 x 10 -4 , 1.8 x 10 -4 and 4.8 x 10 -9 respectively. Calculate the ionisation constant of the corresponding conjugate base. Answer: 1. HF, conjugate base is F K_b = K_w/K_a = \(\frac<1>><6.8>>\) = l.47 x 10 -11 = l.5 x 10 -11

Matter 15. The brand new pH from 0.step one Yards provider away from cyanic acid (HCNO) is 2.34. Calculate the ionization ongoing of your own acidic and its standard of ionization on the service. HCNO \(\rightleftharpoons\) H + + CNO – pH = 2.34 function – record [H + ] = dos.34 otherwise journal [H + ] = – dos.34 = step 3.86 or [H + ] = Antilog 3.86 = cuatro.57 x ten -step 3 Meters [CNO – ] = [H + ] = 4.57 x ten -step three Meters

Question 16. The Ionization constant of nitrous acid is 4.5 x 10 -4 . Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis. Answer: Sodium mtrite is a salt of weak acid, strong base. Hence, K_h = 2.22 x 10_-11 K_w/K_b = 10 -14 /(4.5x 10 -4 )

[OH – ] = ch = 0.04 x 2.thirty six x ten -5 = 944 x 10 -7 pOH = – diary (nine.44 x ten -seven ) = 7 – 0.9750 = six.03 pH = 14 – pOH = 14 – six.03 = seven.97

Question 17. What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298K. For calcium sulphate, K_sp = 9.1 x 10 -6 . Answer: CaSO₄(s) Ca 2 (aq) + SO 2- ₄(aq) If ‘s’ is the solubility of CaSO₄ in moles L – , then K_sp = [Ca 2+ ] x [SO₄ 2- ] = s 2 or

= 3.02 x 10 -3 x 136gL -1 = 0.411 gL -1 (Molar mass of CaSO₄ = 136 g mol -1 ) Thus, for dissolving 0.441 g, water required = I L For dissolving 1g, water required = \(\frac < 1>< 0.411>\)L = 2.43L

The solubility equilibrium regarding over loaded option would be AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) The fresh new solubility regarding AgCl is actually step 1

1. Highlight the distinctions between ionic device and you may solubility tool.
2. The latest solubllity regarding AgCI within the water at the 298 K San Diego escort are step 1.06 x 10 -5 mole each litre. Determine try solubility device at that temperature.

This new solubility harmony from the saturated solution is AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) The solubility of AgCl try step one

1. It is relevant to all or any types of choice.
2. The well worth transform into the change in fraud centration of the ions.

The new solubility harmony about saturated solution is AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) The brand new solubility out of AgCl are 1

1. It’s relevant into the soaked solutions.
2. This has one well worth for a keen electrolyte at the a steady temperature.

2. 06 x 10 -5 mole per litre. [Ag + (aq)] = 1.06 x 10 -5 mol L -1 [Cl – (aq)] = 1.06 x 10 -5 mol L -1 K_sp = [Ag + (aq)] [Cl – (aq)] = (1.06 x 10 -5 mol L -1 ) x (1.06 x 10 -5 mol L -1 ) = 1.12 x 10 -2 moI 2 L -2

Question 19. The value of K of two sparingly soluble salts Ni(OH)₂ and AgCN are 2.0 x 10 -15 and 6 x 10 -17 respectively. Which salt is more soluble? Explain. Answer: