Calculate the hydronium ion concentration and the pH of the solution that result

Question

Calculate the hydronium ion concentration and the pH of the solution that results when 20.0 mL of 0.15M acetic acid, is mixed with 5.0 mL of 0.17 M Sodium Hydroxide. A solution contains 0.10 M iodide ion I^-, and 0.10 M Carbonate ion, CO_3^2- If solid Pb(NO_3)_2 is slowly added to the solution which salt would precipitate first, Pbl_2 or PbCO_3? The enthalpy of vaporization of liquid ethyl ether, (C_2H_5) O_3 is 26.0 kJ/mol at the boiling point of 35.0 deg C. Calculate Delta S degree for a vapor to liquid transformation at 35.0 deg C. Yeast can product ethanol by the fermentation of glucose (C_6H_12O_6), which is the basis for the production of most alcoholic beverages. C_6H_12O_6(aq) rightarrow 2c_2H_5OH(l) + 2CO_2(g)

Explanation / Answer

Q.10: Initial moles of acetic acid = MxV = 0.15 mol/L x 20mL x (1L/1000 mL) = 0.003 mol

Moles of sodium hydroxide(NaOH) added = MxV = 0.17 mol/L x 5 mL x(1L/1000 mL) = 0.00085 mol

After mixing, 0.00085 mol of NaOH will react with 0.00085 mol of CH3COOH to form 0.00085 mol CH3COONa.

The neutralization reaction is

------------- CH3COOH + NaOH ------ > CH3COONa + H2O

init.mol : 0.003 mol, 0.00085 mol, 0 mol

change: - 0.00085 mol, - 0.00085 mol, + 0.00085 mol

eqm.mol: 0.00215 mol, 0 mol, ---------- 0.00085 mol

Now CH3COOH and CH3COONa will act as buffer solution whose pH can be calculated from Hendersen equation

pH = pKa + log[CH3COONa] / [CH3COOH]

=> pH = pKa + log(moles of CH3COONa / moles of CH3COOH) [since V is same for both]

=> pH = 4.74 + log(0.00085 / 0.00215) = 4.34 (answer)

Q.11: The compound with the lowest solubility will precipitate first.

For the dissociation of PbI2

PbI2(s) + H2O ----- > Pb2+(aq) + 2I-(aq) : Ksp = 8.5x10-9

----------------------------- S, ------------ 2S

=> Ksp = 8.5x10-9 = [Pb2+(aq)] x[I-(aq)]2 = S x (2S)2 = 4S3

=> S = cuberoot(8.5x10-9 / 4) = 1.29x10-3 M

For the dissociation of PbCO3

PbCO3(s) + H2O ----- > Pb2+(aq) + CO32-(aq) : Ksp = 7.4x10-14

---------------------------------- S, ------------ S

=> Ksp = 7.4x10-14  = [Pb2+(aq)] x[CO32-(aq)]2 = S x S = S2

=> S = squareroot( 7.4x10-14 ) = 2.72x10-7 M

PbCO3 having lowest solubility will precipitate first (answer)

Q.12: Given deltaHv = 26.0 kJ/mol = 26000 J/mol

boiling point, Tb = 35.0 DegC = 35.0 + 273 = 308 K

Hence deltaS0 = deltaHv / Tb = 26000 J/mol / 308 K = 84.4 Jmol-1K-1 (answer)

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