Question

following reaction the unshaded spheres represent H atoms. Huz CO₃ H₂O H2O a o- H -c 0-0 8) 8) Identify the Bronsted-Lowry acid/base conjugate pairs (1)/(4) and (2)/(3) B) (1)/(3) and (2)/(4) C) (1)/(2) and (3)/(4) 9) Arrange the following 0.10 M aqueous solutions in order of increasing pH: NaOH, HBr, NaCH3CO2, KBr, NH4Br. A) NaOH, NaCH3CO2, KBr, NH4Br, HB B) HBr, KBr, NH Br, NaCH3CO2, NaOH C) NaOH, NaCH3CO2, NH Br, KBr, HB D) HBr, NH4Br, KBr, NaCH3CO2, NaOH 10) 10) What is the magnitude of the change in pH when 0.005 moles of HCl is added to 0.100 L of a buffer solution that is 0.100 M in CH3CO2H and 0.100 M NaCH3CO2? The Ka for acetic acid is 1.8 10-5. A) 0.00 B) 0.20 C) 1.30 D) 0.47 The following pictures represent solutions that contain a weak acid HA (pka = 5.0) and its potassium salt KA. Unshaded spheres represent H atoms, black spheres represent oxygen atoms, and shaded spheres represent A-ions. (K+, H3O+ initially present, OH-initially present and solvent water molecules have been omitted for clarity.) -HA -A 8-H,098-OH 11) 11) Which picture represents the equilibrium state of the solution after addition of one H3O+ ion to the solution shown in picture (1)? A) (2) B) (3) (4) D) (5)

Answer 1

Q-8 ) conjugate acid - base differ by one proton

conjugate base of HCO3^- is CO3^2-

^conjugate acid of H2O is H3O⁺

so answer is B

Q - 9 ) Answer is D since HBr is strong acid , NH4Br is weak acidic salt , KBr is neutral while CH3COONa is basic salt while NaOH is strong base.

Q - 10 ) from Henderson equation

P^H = P^{ka +} log [ CH3COONa ] / [ CH3COOH]

P^{H =} — log ( 1.8 x 10^–5 ) + log ( 0.1 M / 0.1 M ) = 4.74

moles of HCl = 0.005

moles of CH3COONa = moles of CH3COOH =( 0.1 mol / litre ) x 0.1 litre = 0.01 mol

acetate will react with HCl and form CH3COOH and water

final mol of acetate = 0.01 — 0.005 = 0.005 mol

mol of CH3COOH = 0.01 + 0.005 = 0.015

P^H = 4.74 + log ( 0.005 / 0.015 ) = 4.74 —0.47 = 4.27

change in P^H = 4.74 — 4.27 = 0.47

answer is D option

Q -11) answer is C option since initially there are 3 HA & 3 A^— after adding one H⁺ it react with one A^— reducing it to 2 and form one extra HA molecules.so HA molecules are finally 4

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