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- - 144 / Natural Sciences / Chemistry And Polymer Science / Chemi zzes / Buffers en Titrasies | Buffers and Titrations A 0.386 g sample of propanoic acid (CH3CH,COOH, K, = 1.3. x 10-5) was dissolved to a final volume of 50.0 cm, This solution underwent titration with 0.32 M NaOH. Calculate the pH of solution at the equivalence point. %3D fi h- Bi ek Please give the answer to 2 decimal places. Ge pl Answer: Check3. Complete neutralization of 10 ml of phosphoric acid solution by NaOH 0.I N in the presence of phenol phthalein until the appearance of purple color (pH, = 9) 11 ml. of NaOH is consumed. (a) What is the concentration of phosphoric acid? (b) Calculate the indicator error. pk2.1 pK. - 7.2 pK - 1244. Determine the pH of 0.05 M CH3COONa. 5. Determine the pH of 0.05 M NH4NO3. 6. Determine the pH of a solution in which 1.00 mol H2CO3 (Ka = 4.2 x 10-7) and 1.00 mole NaHCO3 is dissolved in enough water to form 1.00 L of solution.
- ng resourc.. (27) Could've Been - [References] Calculate the pH at the halfway point and at the equivalence point for each of the following titrations. a. 100.0 mL of 0.30 M HC7H502 (Ka = 6.4 x 10-5) titrated by 0.10 M NaOH pH at the halfway point = pH at the equivalence point = b. 100.0 mL of 0.40 M C2H;NH2 (K½ = 5.6 × 10¬4) titrated by 0.10 M HNO3 pH at the halfway point = pH at the equivalence point = c. 100.0 mL of 0.70 M HC1 titrated by 0.15 M NaOH pH at the halfway point = pH at the equivalence point =Activity 2. Self-ionization of water and pH, pOH, pKw »1. Write the law of mass action for the reaction of self-ionization of water and calculate the concentrations of [H;O*] and [OH]. H,0(1) + H20(1)= H30*(aq) + OH-(aq) Kw = 1.0x10-14 %3D 2 Since working with yery small powers of 10 can be inconvenient, we commonly use the mathematical3. A solution was prepared by adding 4.95 g sodium acetate (NaC;H;O2) to 250.0 mL of 0.150 M acetic acid (K = 1.75 x 10). Assume negligible volume change. a) What is the pH of the resulting solution? PH: pk2 + log CH3 02) moles saium 4.959 0.06 mol %3D 2cent %3D 82gmu H: -log (1.73x 10) + log 0.24 moles 2cete 250 x 0.15 IL :00375 1000me PH : 4 1.96 O.06 (N2 G Hs0): 0.24m 0.230 L O.0375 : CCHS COOH): 0 15 m Assume 125.0 mL of the solution prepared above was put into a beaker and 82 mg of NaOH was added to it. What is the pH of this new mixture? b)
- 5. What concentration of acetate ion (CH3CO2') in 0.500 M CH3CO2H produces a buffer solution with pH = 5.00? (Ka = 1.8 x 10-5) CH;CO2H (aq) H20 (I) CH;CO, (aq) H30* (aq) + 6. A buffer solution is formed by adding 0.500 mol of sodium acetate (CH3CO2NA) and 0.500 mol of acetic acid (CH;CO2H) to 1.00LH,0. What is the pH of the solution at equilibrium? (K, 1.8 x 10°)8. Given Ka for CH(CH2) COOH - 1.5 x 105 Mat 298 K. Calculate the pHof a. An aqueous solution of 0.10 M CH,(CH:).COOH (Hint: assume (H,O') = [CH3(CH))CO0) b. An aqueous solution of 0.050 M CH,(CH)»COON and 0.050 M CH(CHa);COOH c. 1Lofthe solution in (b) after the addition of 1.0 x 10 molof solid NaOH.4. The pH of 1.50 mol/L acid is measured to be 5.75. For this acid, the Ka is 10-5.25 = [H₂₂0] A. 2.1 x 10-¹2 B. 1.8 x 10-6 C. 2.4 x 10-6 D. very large Chatz (H30] = 1.77x10-6 = 1,8x10-6 In the following information to answer the next question.
- 6. a. Calculate the pH of a buffered solution that is 0.100 M in C6H5CO2H (benzoic acid, Ka = 6.4 x 10-5) and 0.100 M NaC6H5CO2. b. Calculate the pH after 20.0% (by moles) of the benzoic acid is converted to benzoate anion by addition of a strong base. Use the dissociation equilibrium: C6H5CO2H (aq) + H2O (l) ⇌ C6H5CO2- (aq) + H3O+(aq) c. Do the same as in part b, but use the following equilibrium to calculate the pH: C6H5CO2 -(aq) + H2O(l) ⇌C6H5CO2H (aq) + OH- (aq) d. Do your answers in part b and c agree? Explain.3- 13.7. Calculate the ratio of molarities of PO and HPO 2 ions in a buffer solution with a pH of 4 11.0.1. Complete the reaction NaOH → Na+¹ + i 2. What is the pH of a 0.001000 M NaOH solution ? first, find the [H3O+¹] using Kw Kw [ [H3O+¹] = [ |] [ pH = -log( a. acetic c. phosphoric g. sulfurous n. HPO3-2 u. 0.2500 aa. 2.6990 gg. 5.000 x 10-12 mm. conjugate acid b. hydrochloric [] d. perchloric e. hydrofluoric 1. Mg+2 S. HS-1 t. S-2 z. 1x 10-14 x. 1.000 x 10-3 y. 3.0000 CC. 11.0000 dd. 2 ee. 1 ff. 2.000 x 10-³ ii. 0.01250 jj. 0.22100 kk. 0.05656 II. acid pp. conjugate base -2 i. OH-1 j. H30+1 k. SO4² there is complete ionization f. sulfuric h. hydrosulfuric -2 0. H₂PO3-1 p. CO3-² q. HCO3-1 r. H₂S v. 0.1000 w. 0.02500 bb. 1.000 x 10-11 hh. 11.3010 oo. base m. Cl-1