Question

The following pictures represent solutions that contain a weak acid HA (pka = 5.0) and its potassium salt KA. Unshaded spheres represent H atoms and shaded spheres represent A-ions. (K+, H30, OH-, and solvent H2O molecules have been omitted for clarity.) -HA -A IC (1) 12) Which of these solutions are buffers? A) (1) and (2) C) (1), (2) and (3) B) (1) and (3) D) All are buffer solutions. following pictures represent solutions at various stages in the titration of a weak diprotic acid H2A with aqueous KOH. Hashaded spheres represent H atoms, black spheres represent oxygen atoms and shaded spheres represent A2-ions. Ik H3O+ initially pre initially present, OH initially present and solvent water molecules have been omitted for clarity. -HA -HA -A -OH (4) 13) Which picture represents the system halfway between the first and second equivalence points? A) (1) B) (2) C) (3) D) (4) Dhe res points? 13) 14) Calculate the pH of a 0.100 M CH3NH3Cl solution. Kb for methylamine, CH3NH2, is 3.7. 10-4. A) 8.22 B) 222 C) 5.78 D) 11.78 15) Write a balanced equation for the dissociation of the Bronsted-Lowry acid HSO4- in water. A) HSO4- (aq) + H20() - S032-(49) + OH-(ag) B) HSO4-(aq) + H20(1) - H2SO4(0) + OH-(ag) C) HSO4-(aq) H20(1) - SO42-(aq) + H3O+(ag) D) HSO4-(g) + H2O(1) - SO3(g) + OH-(aq) + H2O(1) 16) 16) A buffer solution is prepared by dissolving 27.22 g of KH2PO4 and 3.37 g of KOH in enough water to make 0.100 L of solution. What is the pH of the H2PO4-/HPO42-buffer if the Ka2 = 62.10-87 A) 6.84 C.) 7.21 D) 7.00 B) 7.84

Answer 1

Q - 12 ) answer is D since all options contain A^— and HA

Q - 13 ) answer is A since between first and second equivalence point HA^— and A^2– concentration will be same.

Q - 14 ) hydrolysis of CH3NH3⁺ will take place
K_a ( CH3NH3⁺ ) = K_w / k_b = 10^–14 / 3.7 x 10^–4 = 2.7 x 10^–11

Pka = — log ( 2.7 x 10^–11 ) = 10.57

P^H = 1/2 ( P^ka — log c ) = 5.78

answer is c option

Q -15 ) answer is c it is proton transfer reaction from HSO4^– to H2O

Q - 16 ) moles of KH2PO4 = 27.22 gm / 136.086 gm mol^—1 = 0.2 mol

mol of KOH = 3.37 gm / 56 gm mol^—1 = 0.06 mol

KOH react with H2PO4^- so moles of H2PO4^- = 0.2 —0.06 = 0.14 mol

moles of HPO4^2- = 0.06 mol

P^H = p^Ka2 + log [ HPO4^2- ] / [ H2PO4^- ] = 8 — log 6.2 + log ( 0.06 / 0,14 ) = 6.84

correct answer is a .

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