Brand new pH away from a sample out of white vinegar was step 3

Brand new pH away from a sample out of white vinegar was step 3

Question 13. 76. Calculate the concentration of hydrogen ion in it. Answer: pH = – logten [H3O + ] = – log10 = – pH = – 3.76 = \(\overline\).24 [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 Kb = Kw/Ka = \(\frac<1>><6.8>>\) = l.47 x 10 -11 = l.5 x 10 -11

Question fifteen. Brand new pH off 0.step 1 Yards service out of cyanic acid (HCNO) was 2.34. Determine new ionization ongoing of one’s acidic and its own degree of ionization throughout the provider. HCNO \(\rightleftharpoons\) H + + CNO – pH = dos.34 setting – diary [H + ] = dos.34 or record [H + ] = – 2.34 = 3.86 otherwise [H + ] = Antilog 3.86 = cuatro.57 x ten -step 3 M [CNO – ] = [H + ] = 4.57 x 10 -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, Kh = 2.22 x 10-11 Kw/Kb = 10 -14 /(4.5x 10 -4 )

[OH – ] = ch = 0.04 x dos.36 x 10 -5 = 944 x 10 -seven pOH = – log (nine.forty-two x ten -eight ) = 7 – 0.9750 = six.03 pH = fourteen – pOH = 14 – 6.03 = seven.97

Question 17. What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298K. For calcium sulphate, Ksp = 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 Ksp = [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 latest solubility balance regarding the over loaded solution is AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) The new solubility regarding AgCl is actually step 1

  1. Point out the distinctions anywhere between ionic equipment and you will solubility tool.
  2. The newest solubllity regarding AgCI in water during the 298 K was step one.06 x 10 -5 mole for every litre. Determine is solubility unit at this temperature.

The fresh solubility harmony from the over loaded solution is AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) The new solubility https://datingranking.net/escort-directory/kent/ out-of AgCl is step one

  1. It is appropriate to all the type of solutions.
  2. Their worth transform on improvement in scam centration of one’s ions.

The newest solubility balance throughout the over loaded solution is AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) This new solubility out-of AgCl try 1

  1. It is relevant towards soaked selection.
  2. It has a definite worth to have a keen electrolyte during the a stable temperatures.

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 Ksp = [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: