high-point escort

24 Juin The fresh new pH out-of a sample out of white vinegar is 3

Question 13. 76. Calculate the concentration of hydrogen ion in it. Answer: pH = – log_ten [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 K_b = K_w/K_a = \(\frac<1>><6.8>>\) = l.47 x 10 -eleven = l.5 x 10 -11

Matter fifteen. New pH off 0.step 1 Meters service off cyanic acid (HCNO) is dos.34. Determine the fresh ionization constant of acid and its particular level of ionization regarding the service. HCNO \(\rightleftharpoons\) H + + CNO – pH = dos.34 mode – record [H + ] = 2.34 or journal [H + ] = – 2.34 = 3.86 otherwise [H + ] = Antilog step three.86 = 4.57 x 10 -3 M [CNO – ] = [H + ] = cuatro.57 x 10 -step three Yards

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.thirty-six x 10 -5 = 944 x 10 -seven pOH = – record (9.forty two x 10 -seven ) = eight – 0.9750 = six.03 pH = fourteen – pOH = 14 – six.03 = 7.97