The pH off a sample out of white vinegar try 3

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 new pH regarding 0.1 Meters service out-of cyanic acidic (HCNO) are 2.34. Assess the fresh new ionization lingering of your own acidic and its particular standard of ionization from the provider. HCNO \(\rightleftharpoons\) H + + CNO – pH = dos.34 setting – journal [H + ] = dos.34 or journal [H + ] = – 2.34 = step 3.86 or [H + ] = Antilog step three.86 = 4.57 x ten -step three Meters [CNO – ] = [H + ] = 4.57 x 10 -3 M

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 dos.36 x 10 -5 = www.datingranking.net/escort-directory/rancho-cucamonga/ 944 x 10 -seven pOH = – diary (9.44 x 10 -seven ) = eight – 0.9750 = six.03 pH = 14 – pOH = fourteen – six.03 = eight.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 fresh new solubility equilibrium from the over loaded option would be AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) This new solubility off AgCl is step one

1. Highlight the differences between ionic tool and you will solubility tool.
2. The solubllity off AgCI in water at 298 K try step 1.06 x ten -5 mole for every litre. Determine is solubility tool at that temperatures.

The brand new solubility equilibrium on the over loaded solution is AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) The latest solubility regarding AgCl is step one

1. It is appropriate to all or any sort of alternatives.
2. Their really worth alter into the change in ripoff centration of the ions.

The brand new solubility harmony on saturated option would be AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) New solubility out of AgCl is 1

1. It is applicable for the saturated selection.
2. It has a particular worth to possess an electrolyte during the a stable 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: