Solutions for Simple ph and concentration calculation problems

Transcription

1 Revision: Solutions for Simple ph and concentration calculation problems 1. hat is the mass of a 51.6 ml sample of gasoline, which has a density of 0.70 g/cm? ρ (density) = m (mass), from this m = ρ V = 0.70 g/cm 51.6 cm = 6.12 g V (volume) 2. A flask contains 25.0 ml of ether weighing g. hat is the density of ether? m g ρ = = V 25 cm = g/cm. How many moles of sodium chloride should be put in a 50.0 ml volumetric flask to give a 0.15 M NaCl solution when the flask is filled with water? How many grams of NaCl is this? (NaCl: M.. = 58.5 g/mol) n (molar amount) c (molar concentration or molarity) =, from this n = c V = V (volume) 0.15 mmol/cm 50 cm = 7.5 mmol = mol. 1 mol of NaCl is 58.5 g, that is moles are: m = n M.. (molar weight) = mol 58.5 g/mol = g 4. A sample of NaOH weighing 0.8 g is placed in a 50.0 ml volumetric flask. The flask is then filled with water to the mark. hat is the molarity of the resulting solution? Is it necessary to use a volumetric flask to make this solution? (NaOH: M.. = 40.0 g/mol) To calculate the molarity (c) of the solution, we need the molar amount of the solute and the volume of the solution. The volume (50 ml) is given. To calculate the molar amount, we use the following equation: m = n M.., that is n = m 0.8 g = = mol = 9.5 mmol. M.. 40 g/mol n 9.5 mmol Than: c = = = 0.19 M V 50 cm It is not necessary to use a volumetric flask, since, anyway, the concentration cannot be very accurate because of the carbonate content of the solid NaOH. 5. How many moles of NaOH are contained in 1 ml of 0.15 M NaOH? c = V n, that is n = c V = 0.15 mmol/cm 1 cm = 4.65 mmol = mol. 6. How would you prepare 425 g of an aqueous solution containing 2.40% (by mass) of sodium acetate?

2 The mass of sodium acetate needed is 2.40 % of the total mass that is 425 g 2.40 = 10.2 g of sodium acetate. The amount of water needed is the remaining 100 mass that is 425 g 10.2 g = g of water. 7. You are given 5.00 ml of 14.8 M NH solution. hat will the final volume be after this solution is diluted with water to give 1.00 M NH solution? To calculate the final volume of the solution, we need to know the molar amount of the solute and the concentration of the final solution. The final concentration (1.00 M) is given. The molar amount of NH is the same in the initial 5 ml and in the final solution, because we do not add any more NH during the dilution. Tis molar amount (calculted from the initial solution) is: n = c V = 14.8 mmol/cm 5 cm = 74 mmol = mol NH Now, we know the concentration and also the molar amount of NH in the final solution, so we can calculate the volume of it: n n 74 mmol c = that is V = = = 74 cm = 74 ml V c 1mmol/cm 8. Commercially available concentrated hydrocloric acid is an aqueous solution containing 8% HCl by mass. It s density is 1.19 g/ml. hat is the molarity of this solution? How many milliliters of this concentrated solution is required to make 1.00 L of 0.10 M HCl solution? To calculate the molarity (c) of the solution, we need the molar amount of the solute (HCl) and the volume of the solution. Let s take 1 ml of this solution (you can choose also a different volume). In this case, from the density of the solution, we can calculate the mass of the solution, which is: m = ρ V = 1.19 g/ml 1 ml = 1.19 g 8 8 % of this mass is HCl, that is the mass of HCl (solute) is 1.19 g = g. So, the molar amount of HCl is: m g n = = = mol = 12.4 mmol. M g/mol Finally, the concentration of this solution is: n 12.4 mmol c = = = 12.4 M V 1cm If we want to make 1 L of 0.10 M HCl soluton, than we need the volumewhich contains the required amount of HCl. In the final solution we need: n = c V = 0.1 M 1 L = 0.1 mol of HCl. The same amount is needed in the concentrated solution, that is (for the concentrated solution): n 0.1 mol V = = = L = 8.1 ml of concentrated HCl solution. c 12.4 mol/l

3 Simple ph calculations: 1. Calculate the ph of the solution: a) [H O + ] = 0.01 mol/dm b) [H O + ] = mol/dm c) [H O + ] = mol/dm a) ph = 2.00 b) ph = 2.9 c) ph = Calculate the [H O + ] and the ] concentrations in the following solutions: a) ph = 2.2 b) ph = 6.45 c) ph = 11.2 a) [H O + ] = mol/dm ] = mol/dm b) [H O + ] = mol/dm ] = mol/dm c) [H O + ] = mol/dm ] = mol/dm. A sample of orange juice has an equilibrium hydrogen ion concentration of M. hat is the ph? Is the solution acidic, basic or neutral? ph = -log[h + ] = -log( ) =.54 ph < 7.00 so the solution is acidic. 4. A saturated solution of potassium hydroxide has an equilibrium hydroxide ion concentration of 0.05 M. hat is the ph? Is the solution acidic, basic or neutral? poh = -log - ] = -log(0.05) = 1.0 ph + poh = that is ph = poh = = ph > 7.00 so the solution is basic. 5. A HClO 4 solution has an analytical concentration of M. hat is the equilibrium hydrogen ion concentration and the ph? In the case of strong acids, the dissociation is 100 %, so the concentration of the acid and the H + ion is the same (expect when c strong acid < 10 6 M, but this is not our case). So, [H + ] = c strong acid = M, and ph = -log[h + ] = -log(0.012) = A NaOH solution has an analytical concentration of 10-7 M. hat is the equilibrium hydroxide ion concentration, the poh and the ph? In the case of strong bases, the dissociation is 100 %, but now c strong base < 10 6 M, so the selfdissociation of the water has also to be taken into account: K ] = cstrong base + ] ] = 10 ] ] = ] ] + 10 ] = 0

4 - ( 10 ] = 7 ) ± ( 10 7 ) ( ) = M 2 1 poh = -log - ] = -log( ) = 6.79 ph + poh = that is ph = poh = = Calculate the ph of pure water at 25.0 and 50.0 C (K = and , respectively) In pure water there is no extra H + or OH - so [H + ] = - ]. At the same time so [H + ] - ] = K. From these two equations: [H + ] = K At 25 C, [H + ] = At 50 C, [H + ] = 6.65 K = K = = , so ph = -log( ) = = , so ph = -log( ) = ml of HCl of unknown concentration is titrated with potassium hydroxide whose concentration is M. Calculate the ph at: a) 0 %; b) 40 %; c) 100 %; d) 160 % degree of titration. Up to the equivalence point ml of KOH solution is consumed. a) 0% n(naoh) = c V = = mol n(naoh) = n(hcl) = mol c(hcl) = / = M ph = -log[h + ] ph = b) 40% 40% of HCl is titrated, 60% remains. n(hcl) = = mol V(sol) = = dm n(naoh) = = mol V(NaOH solution) = dm c(hcl) = / = M ph = -log[h + ] ph = 1.70 c) 100 % ph = d) 160 % V(solution) = V(HCl sample) + V(NaOH eqiuv.) + V(NaOH extra)

5 V(solution) = = dm n(naoh) = mol c(naoh) = / = M poh = -log ] poh = -log ( ) poh = ph = = ml of HCl of mol/dm concentration is titrated with sodium hydroxide whose concentration is mol/dm. Calculate the ph at: a) 0 %; b) 20 %; c) 50 %; d) 120 % degree of titration. a) 0 % c(hcl) = mol/dm ph = - log [H + ] ph = - log 0.1 ph = b) 20 % n(hcl) = = mol The amount of NaOH needed for the neutralization of 20 % of the acid: n(naoh) = = mol V(NaOH) = = dm V(solution) = V(HCl sample) + V(NaOH added to the system) V(solution) = = dm n( HCl) c(hcl) = = = mol/dm V ( sol) ph = - log[h + ] ph = - log ph = c) 50 % n(hcl) = = mol The amount of NaOH needed for the neutralization of 50 % of the acid: n(naoh) = = mol V(NaOH) = = dm 0.2 V(solution) = = dm

6 n( HCl) c(hcl) = = = mol/dm V ( sol) ph = - log[h + ] ph = - log 0.04 ph = 1.98 d) 120 % After the equivalence point the ph of the solution is determined by the excess of NaOH. n(naoh) = = mol n(naoh) excess = = mol V(NaOH) = = dm V(solution) = = 0.02 dm n( NaOH ) c(naoh) = = = mol/dm V ( NaOH ) 0.02 poh = - log ] poh = - log poh = 1.90 ph = 14 ph = =