Calculate rate of reactions A + B rightarrow product a 38 b .278 c 13.2 d 42.0 1

Question

Calculate rate of reactions A + B rightarrow product a 38 b .278 c 13.2 d 42.0 1 What is PH of sol^n that is made by dissolving 6.78 g of barium hydroxide to total volume of 425.0 mL 2 a 13 b 12.8 c 13.270 d 12.500 2 What is the ph of Duffer sol^n that is .15 M ammonium cheoride & .18 M ammonia if kq = 5.6 times 10^-10 2 a 9.33 b 4.66 c 9.18 d 4.82 3 What is kb of weak base if .15 M sol^n B_(qq) + H_2O BH^+_(eq) + OH^-_(qq) a 3.3 times 10^- 11 b 4.8 times 10^- 8 c 2.7 times 10^- 5 d 8.6 times 10^- 9.

Explanation / Answer

1) Molar mass of Ba(OH)2 = 171.34 g/mole

Thus, moles of Ba(OH)2 in 6.78 g of it = mass/molar mass = 6.78/171.34 = 0.04

Thus, [Ba(OH)2] = moles of Ba(OH)2/volume of solution in litres = 0.04/0.425 = 0.093 M

Thus, [OH-] = 2*[Ba(OH)2] = 0.186 M

pOH = -log[OH-] = 0.73

Hence pH = 14 - pOH = 13.27

The correct option is :- (c)

2) Using Henderson's equation :-

pH = pKa + log{[NH4Cl]/[NH3]}

pKa = -logKa = 9.252

Thus, pH = 9.252 + log(0.15/0.18) = 9.173

Thus, the correct option is :- (c)

3) [H+] = 10-pH = 5.012*10-12 M

Thus, [OH-] = 10-14/[H+] = 1.995*10-3 M

Now, Kb = [OH-]*[BH+]/[B] & [BH+] = [OH-]

Thus, Kb = (1.995*10-3)2/(0.15-0.001995) = 2.7*10-5

Thus, the correct option is :- (c)

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