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- If a 0.3 M solution of NaOH was used to titrate 100 mL of 0.3 M solution of CH3COOH (Ka =1.8 x 10 -5) and the following volumes of NaOH were recorded: 102, 97, 99, 98, 101, 106 mL. The pH Value is………………when added 99.0 mL of 0.3 M NaOH The pH of the solution at the equivalence point more than 7 Because................................... Calculate the 99% confidence limits of the mean? And use them to decide whether there is any evidence of systematic error? [ hints: For N=6 and the 99 % confidence interval, the value of t is 4.03].Calculate the pH at the following points for the titration of 50.00 mL of 0.0100 mol L-1 H2SO4 with 0.0100 mol L-1 NaOH standard solution.Vbase = 0.0; 10.0; 20.0; 25.0; 25.5; 40.0; 50.0 and 100.0.Plot a graph of pH versus Vbase.Perform the same procedure for H2SO4 and 0.01 mol L-1 NaOH concentrations.Generate the titration curves for the two solutions, compare them and justify.Be sure to answer all parts. Find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH3CH,);N (K, = 5.2 x 10-4), with 0.1000 M HCl solution after the following additions of titrant. (a) 13.00 mL: pH = 10.305 (b) 20.10 mL: pH = (c) 29.00 mL: pH = %3!
- The value of Ksp for Mg3(AsO4)2 is 2.1 x 10-20. TheAsO43 - ion is derived from the weak acid H3AsO4 (pKa1 =2.22; pKa2 = 6.98; pKa3 = 11.50). (a) Calculate the molarsolubility of Mg3(AsO4)2 in water. (b) Calculate the pH of asaturated solution of Mg3(AsO4)2 in water.Consider the titration of 100.0 mL of 0.100 M hydrazine (H2NNH2, K, = 3.0×10-6) by 0.200 M HNO3. Despite the symmetric nature of the hydrazine's molecular structure, it accepts only a single proton from HNO3 in aqueous solution. (Kw = 1.00×10-14) %3D %3D (a) Calculate the pH of the solution after addition of 0.0, 20.0, 25.0, 40.0, 50.0, and 100.0 mL of HNO3 to the solution. (Note: you need to consider the range of applicability of the Henderson- Hasselbalch equation and need to check if water autoionization can be neglected. See Figure 15.18 in the textbook.) (b) Based on your answer to part (a), sketch the titration curve, marking the calculated points and labeling the equivalence and half-equivalence points. (c) Although still weak, ammonia (NH3, Kb = 1.8×10-5) is a stronger base than hydrazine. Sketch the titration curve for 100.0 mL of 0.100 M ammonia by 0.200 M HNO3 on the same axes as your answer to part (b); be sure to label the curve. %3D Note: a qualitative sketch is…3. 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 - 124
- If the standard solutions had unknowingly been made up to be 0.0024 M AgNO3 and 0.0040 1 MK2CrO4, what would be the value of Ksp? The value of Ksp would be larger. The value of Ksp would be smaller. The value of Ksp would be the same. O More information is needed to answer this question.5. In the titration of 15.00 mL of 0.200 M of a weak base, B:, (Kb = 7.25 x 10³) with 0.100 M HCI. calculate the pH after the addition of the following volumes of titrant (mL); (a) 0.00 ; (b) 10.00: (c ) 15.00 ; (d) 20.00 mL. %3D9-20. (a) Calculate how many milliliters of 0.100 M HCI should be added to how many grams of sodium acetate dihy- drate (NaOAc· 2H2O, FM 118.06) to prepare 250.0 mL of 0.100 M buffer, pH 5.00. (b) If you mixed what you calculated, the pH would not be 5.00. Describe how you would actually prepare this buffer in the lab.
- Determine the pH during the titration of 39.6 mL of 0.226 M trimethylamine ((CH3)3N, Kp = 6.3x10-5) by 0.226 M HNO3 at the following points. (Assume the titration is done at %3D 25 °C.) Note that state symbols are not shown for species in this problem. (a) Before the addition of any HNO3 (b) After the addition of 15.8 mL of HNO3 (c) At the titration midpoint (d) At the equivalence point (e) After adding 57.8 mL of HNO3Consider the titration of 40.0 mL of 0.0600 M HONH2 (a weak base; Kb = 1.10e-08) with 0.100 M HI. Calculate the pH after the following volumes of titrant have been added: (d) 18.0 mLpH = (e) 24.0 mLpH = (f) 38.4 mLpH =3. 40.00 mL of 0.0900 M NaOH is diluted to 100mL and titrated with 0.1000M HCI. Calculate the pH after the addition of the following volumes of titrant (mL); (a) 0.00; (b) 10.00 ; (c) 18.00: (d) 30.00 ; (e) 35.95 ; (f) 36.00 (g) 36.05 ; (h) 40.00.
