Your shopping cart is empty!
Explore Related Concepts


help with chemistry back titration calculations
Best Results From Yahoo Answers Youtube
From Yahoo Answers
Question:2. Sodium carbonate exists in hydrated form, Na2CO3.xH2O, in the solid state. 3.5 g of a sodium carbonate sample was dissolved in water and the volume made up to 250 cm3. 25.0 cm3 of this solution was titrated against 0.1 moldm3 HCl and 24.5 cm3 of the acid were required. Calculate the value of x given the equation:
Na2CO3 + 2HCl 2NaCl + CO2 + H2O
Answers:moles HCl = 0.0245 dm^3 x 0.1 mol/dm^3 = 0.00245 moles Na2CO3 in 25.0 cm^3 = 0.00245/2=0.001225 moles Na2CO3 in 250 cm^3 = 0.01225 mass Na2CO3 = 0.01225 mol x 105.9876 g/mol=1.30 g mass water = 3.5  1.30=2.2 g moles water = 2.2 g / 18.02 g/mol=0.122 0.122 / 0.01225 = 10 x = 10
Answers:moles HCl = 0.0245 dm^3 x 0.1 mol/dm^3 = 0.00245 moles Na2CO3 in 25.0 cm^3 = 0.00245/2=0.001225 moles Na2CO3 in 250 cm^3 = 0.01225 mass Na2CO3 = 0.01225 mol x 105.9876 g/mol=1.30 g mass water = 3.5  1.30=2.2 g moles water = 2.2 g / 18.02 g/mol=0.122 0.122 / 0.01225 = 10 x = 10
Question:a .5130g aspirin tablet which was dissolved in 95% ethanol required 27.98mL of 0.1M NaOH for neutralization. Then an additional 42.78mL of 0.1M NaOH was added and the solution was heated to hydrolyze the acetylsalicylic acid. after the reaction mixture was cooled, the excess base was backtitrated with 14.29mL of 0.1056M HCl. how many grams of acetylsalicylic acid are in the tablet? what % w/w of the tablet is acetylsalicylic acid?
Answers:CH3COOC6H4COOH (s) + 2NaOH (aq) > CH3COONa (aq) + HOC6H4COONa (aq) + H2O (l) NaOH (aq) + HCl (aq) > NaCl (aq) + H2O (l) Number of mols HCl = Concentration * Volume in L Number of mols HCl = 0.1056 * 0.01429 = 1.509 * 10^3 moles Number of mols HCl = Number of mols NaOH in excess while backtitrating. Number of mols NaOH given (before hydrolysis) = Concentration * Volume in L Number of mols NaOH given = 0.1000 * 0.04278 = 4.429 * 10^3moles Number of mols of NaOH used in hydrolysis of Aspirin = Number of mols of NaOH given Number of mols of NaOH in excess Number of mols of NaOH which reacted = 0.004429  0.001509 Number of mols of NaOH which reacted = 0.002769 = 2.769 * 10^3 moles. by mole ratio (from eq. 1) we have.. Number of mols of Aspirin hydrolysed = 1/2 * Number of mols of NaOH (which reacted with the aspirin.. which is what we calculated above.) = 0.5 *(2.769 * 10^3) moles = 1.3845 *10^3 moles. Molar mass of Aspirin (CH3COOC6H4COOH ) = C (9*12) + H (8*1) + O (4*16) Mm of Aspirin (C9H8O4) = 180g Mass of Aspirin = Number of moles * Mm Mass of Aspirin = 1.3845 *10^3 moles * 180g = 0.24921g w/w % of pure Aspirin = (mass of aspirin used/mass of tablet)*100 w/w % of pure Aspirin = (0.24921/0.5310)*100 w/w % of pure Aspirin = 48.57% I am in UTM too.. This is the way I solved this problem.. And I guess it should be right.. :).. Take care..
Answers:CH3COOC6H4COOH (s) + 2NaOH (aq) > CH3COONa (aq) + HOC6H4COONa (aq) + H2O (l) NaOH (aq) + HCl (aq) > NaCl (aq) + H2O (l) Number of mols HCl = Concentration * Volume in L Number of mols HCl = 0.1056 * 0.01429 = 1.509 * 10^3 moles Number of mols HCl = Number of mols NaOH in excess while backtitrating. Number of mols NaOH given (before hydrolysis) = Concentration * Volume in L Number of mols NaOH given = 0.1000 * 0.04278 = 4.429 * 10^3moles Number of mols of NaOH used in hydrolysis of Aspirin = Number of mols of NaOH given Number of mols of NaOH in excess Number of mols of NaOH which reacted = 0.004429  0.001509 Number of mols of NaOH which reacted = 0.002769 = 2.769 * 10^3 moles. by mole ratio (from eq. 1) we have.. Number of mols of Aspirin hydrolysed = 1/2 * Number of mols of NaOH (which reacted with the aspirin.. which is what we calculated above.) = 0.5 *(2.769 * 10^3) moles = 1.3845 *10^3 moles. Molar mass of Aspirin (CH3COOC6H4COOH ) = C (9*12) + H (8*1) + O (4*16) Mm of Aspirin (C9H8O4) = 180g Mass of Aspirin = Number of moles * Mm Mass of Aspirin = 1.3845 *10^3 moles * 180g = 0.24921g w/w % of pure Aspirin = (mass of aspirin used/mass of tablet)*100 w/w % of pure Aspirin = (0.24921/0.5310)*100 w/w % of pure Aspirin = 48.57% I am in UTM too.. This is the way I solved this problem.. And I guess it should be right.. :).. Take care..
Question:representing the unknown acid as HA, the equation for the reaction with sodium hydroxide is shown below:
HA + NaOH = NaA + H20
a.) use the average titre to calculate the concentration in mol dm3 of the diluted acid
b.) use your result from a.) to determine the concentration in mol dm3 of the original acid solution
c.) the manufacturer claimed that the concentration of the acid in the original solution of the cleaner was 100g dm3. Use your result from part c.) to predict the Mr of the unknown acid.
average titre = 21.90cm3
much helped appreciated i have a test tomorrow :s basically questions 3,4 and 5 from this question paper http://store.aqa.org.uk/qual/gce/pdf/AQACHM3TWSQP07.PDF x
Answers:21.9ml = titre HA + NaOH > NaA + HOH, everything is a 1:1 ratio you used 21.9ml to neutralize what concentration of NaOH? without that, we cannot determine the molarity of the HA.
Answers:21.9ml = titre HA + NaOH > NaA + HOH, everything is a 1:1 ratio you used 21.9ml to neutralize what concentration of NaOH? without that, we cannot determine the molarity of the HA.
Question:1. What is the chemical definition of a buffer solution? Give an example.
2. The (Ka) of benzoic acid, C7H5O2H, is 6.5 x 10(5). What is its buffer range?
3. How many g of sodium benzoate, NaC7H5O2, should be mixed with 3.00 moles of benzoic acid, C7H5O2H, (Ka)=6.5 x 10(5) in a 5.0L to make a solution that has a pHof 1.20?
4. Calculate the pH of a solution prepared by mixing 30.00 mL of a 0.15M KOH solution with 50.00 mL of a .39M HBr solution.
5. 45.0 mL of a 0.10M formic acid (HCHO2, (Ka)= 1.7 x 104) solution was titrated with a 0.20M NaOH solution. What is the equivalent volume?
6. What is the pH of the acidic solution before titration?
7. How is the solubility of CaF2 in water different from its solubility in Ca(NO3)2? Is CaF2 more or less soluble in water than Ca(NO3)2?
8. Calculate the solubility of CaF2 in a 0.15 M solution of Ca(NO3)2.
Answers:1. A buffer solution is a solution in which there is very little change in pH when a small quantity of acid or base is added to it. It is commonly made by maxing a weak acid and one of its salts or a weak base and one of its salts into an aqueous solution. Example: A mixture of ethanoic acid and sodium ethanoate in solution. 2) Ka = 6.5 x 10^5 pKa = log(Ka) = log(6.5 x 10^5) = 4.187 Buffer range = pKa  1 to pKa + 1 = 4.187  1 to 4.187 + 1 = 3.187 to 5.187 = 3.2 to 5.2 (after rounding) Ans: 3.2 to 5.2 3)[acid] = 3.00 moles/5.0 L = 0.60 M pH = pKa + log([salt]/[acid]) 1.20 = log(6.5 x 10^5) + log([salt]/0.60) 1.20 = 4.187 + log[salt]  log(0.60) 1.20 = 4.187 + log[salt]  (0.222) 1.20 = 4.187 + log[salt] + 0.222 log[salt] = 1.20  4.187  0.222 = 3.209 [salt] = 10^3.209 = 6.18 x 10^4 M Number of moles of the salt in 5.0 L solution = 6.18 x 10^4 x 5.0 = 3.09 x 10^3 Molar mass of the salt (NaC7H5O2) = 23 + 12x7 + 1x5 + 16x2 = 144 g/mol Mass of the salt = moles x g/mol = 3.09 x 10^3 x 144 g = 0.445 g Ans: 0.445 g 4) Moles of KOH = 30.00/1000 L x 0.15 mol/L = 0.0045 mol Moles of HBr = 50.00/1000 L x 0.39 mol/L = 0.0195 mol 0.0045 mol KOH reacts with 0.0045 mol HBr. Remaining moles of HBr = 0.0195  0.0045 = 0.015 Volume of the solution = 30.00 + 50.00 = 80.00 mL = 0.08 L [HBr] = 0.015/0.08 = 0.1875 HBr is a strong acod. Therefore [H+] = [HBr] = 0.1875 pH = log[H+] = log(0.1875) = 0.727 Ans: 0.727 (Note: You can get more accurate result by using Ka for HBr. I did not use because it is not given in the problem.) 5) 1 mol HCHO2 reacts with 1 mol NaOH Therefore M(formic acid) x V(formic acid) = M(NaOH) x V(NaOH) 0.10 x 45.0 = 0.20 x V(NaOH) V(NaOH) = 0.10 x 45.0/0.20 = 22.5 mL Ans: 22.5 mL 6) HCHO2 <> H+ + CHO2  .......0.10..................0............0 (initially) .........x.....................x............x (reaction) ......0.10x................x............x (equilibrium) (x)(x)/(0.10x) = Ka = 1.7 x 10^4 Formic acid is a weak acid. Therefore x << 0.10 x^2/0.10 = 1.7 x 10^4 x^2 = 1.7 x 10^5 [H+] = x = sqrt(1.7 x 10^5) = 0.0041231 pH = log[H+] = log(0.0041231) = 2.38 Ans: 2.38 7)CaF2 and Ca(NO3)2 have Ca 2+ as a common ion. Due to common ion effect, the solubility of CaF2 is reduced in Ca(NO3)2. Thus CaF2 is more soluble in water than in Ca(NO3)2. 8) Ksp of CaF2 = 3.9 x 10^11 CaF2 <====> Ca(2+) + 2F If X mol of CaF2 dissolves then X mol of Ca(2+) is produced and 2X mol of F is produced. Calcium nitrate provides 0.15 M Ca 2+ concentration. [Ca 2+] = 0.15 + X Ksp = 3.9 x 10^11 = [X+.15]*[2X]^2 Since .15 >> X, therefore X+.15 is very close to .15 Ksp = 3.9 x 10^11 = [.15]*[2X]*2 3.9 x 10^11 = .6 x^2 x = sqrt(3.9 x 10^11/.6) = 8.06 x 10^6 Ans: 8.06 x 10^6 __________.
Answers:1. A buffer solution is a solution in which there is very little change in pH when a small quantity of acid or base is added to it. It is commonly made by maxing a weak acid and one of its salts or a weak base and one of its salts into an aqueous solution. Example: A mixture of ethanoic acid and sodium ethanoate in solution. 2) Ka = 6.5 x 10^5 pKa = log(Ka) = log(6.5 x 10^5) = 4.187 Buffer range = pKa  1 to pKa + 1 = 4.187  1 to 4.187 + 1 = 3.187 to 5.187 = 3.2 to 5.2 (after rounding) Ans: 3.2 to 5.2 3)[acid] = 3.00 moles/5.0 L = 0.60 M pH = pKa + log([salt]/[acid]) 1.20 = log(6.5 x 10^5) + log([salt]/0.60) 1.20 = 4.187 + log[salt]  log(0.60) 1.20 = 4.187 + log[salt]  (0.222) 1.20 = 4.187 + log[salt] + 0.222 log[salt] = 1.20  4.187  0.222 = 3.209 [salt] = 10^3.209 = 6.18 x 10^4 M Number of moles of the salt in 5.0 L solution = 6.18 x 10^4 x 5.0 = 3.09 x 10^3 Molar mass of the salt (NaC7H5O2) = 23 + 12x7 + 1x5 + 16x2 = 144 g/mol Mass of the salt = moles x g/mol = 3.09 x 10^3 x 144 g = 0.445 g Ans: 0.445 g 4) Moles of KOH = 30.00/1000 L x 0.15 mol/L = 0.0045 mol Moles of HBr = 50.00/1000 L x 0.39 mol/L = 0.0195 mol 0.0045 mol KOH reacts with 0.0045 mol HBr. Remaining moles of HBr = 0.0195  0.0045 = 0.015 Volume of the solution = 30.00 + 50.00 = 80.00 mL = 0.08 L [HBr] = 0.015/0.08 = 0.1875 HBr is a strong acod. Therefore [H+] = [HBr] = 0.1875 pH = log[H+] = log(0.1875) = 0.727 Ans: 0.727 (Note: You can get more accurate result by using Ka for HBr. I did not use because it is not given in the problem.) 5) 1 mol HCHO2 reacts with 1 mol NaOH Therefore M(formic acid) x V(formic acid) = M(NaOH) x V(NaOH) 0.10 x 45.0 = 0.20 x V(NaOH) V(NaOH) = 0.10 x 45.0/0.20 = 22.5 mL Ans: 22.5 mL 6) HCHO2 <> H+ + CHO2  .......0.10..................0............0 (initially) .........x.....................x............x (reaction) ......0.10x................x............x (equilibrium) (x)(x)/(0.10x) = Ka = 1.7 x 10^4 Formic acid is a weak acid. Therefore x << 0.10 x^2/0.10 = 1.7 x 10^4 x^2 = 1.7 x 10^5 [H+] = x = sqrt(1.7 x 10^5) = 0.0041231 pH = log[H+] = log(0.0041231) = 2.38 Ans: 2.38 7)CaF2 and Ca(NO3)2 have Ca 2+ as a common ion. Due to common ion effect, the solubility of CaF2 is reduced in Ca(NO3)2. Thus CaF2 is more soluble in water than in Ca(NO3)2. 8) Ksp of CaF2 = 3.9 x 10^11 CaF2 <====> Ca(2+) + 2F If X mol of CaF2 dissolves then X mol of Ca(2+) is produced and 2X mol of F is produced. Calcium nitrate provides 0.15 M Ca 2+ concentration. [Ca 2+] = 0.15 + X Ksp = 3.9 x 10^11 = [X+.15]*[2X]^2 Since .15 >> X, therefore X+.15 is very close to .15 Ksp = 3.9 x 10^11 = [.15]*[2X]*2 3.9 x 10^11 = .6 x^2 x = sqrt(3.9 x 10^11/.6) = 8.06 x 10^6 Ans: 8.06 x 10^6 __________.
From Youtube
3. Titration Calculations, Sodium Carbonate :Worked example of a titration question involving the neutralisation of sodium carbonate by hydrochloric acid. The unknown concentration of hydrochloric acid is calculated. Suitable for A2 chemistry or equivalent.
4. Titration Calculations, Dichromate with Iodide :Worked example of a titration question involving the oxidation of iodide by dichromate, followed by the reaction of iodine with sodium thiosulphate. The unknown concentration of thiosulphate is calculated. Suitable for A2 chemistry or equivalent.