Please calculate the pH at the equivalence point when 25 ml of 0.10 NH3 is titrated with 25 mL of a 0.10 M HCl solution (KaNH4+ is 5.6 X 10-10).
25 mL of 0.10 M ammonia solution completely neutralsies 25 mL of 0.10 M HCl solution to form ammonium chloride
Ammonium chloride provides ammonium ions
Let x M be the change in to reach equilibrium. Prepare an ICE table
Initial concentration (M) | 0.050 | 0.00 | 0.00 |
Change in concentration (M) | - x | + x | + x |
Equilibrium concentration (M) | 0.050 - x | x | x |
The acid dissociation constant
Since acid dissociation constant is very small, approximate 0.050-x to 0.050
Please calculate the pH at the equivalence point when 25 ml of 0.10 NH3 is...
calculate the pH of the solution at the equivalence point when 25.0 mL of .10 M benzoic acid is titrated with 0.10 M potassium hydroxide. The Ka for benzoic acid is 6.3 x 10^-5
Calculate the pH at the equivalence point when 25.0mL of 0.100M NH3 is titrated with 0.100M HCl. The reaction is given as follows : NH3 + H30+ -> NH4+ + H20 Show complete solution . Will rate your answer. Draw titration curve and label first and second equivalence point.
Calculate the pH at the equivalence point when 25.0mL of 0.100M NH3 is titrated with 0.100M HCl. The reaction is given as follows : NH3 + H30+ -> NH4+ + H20 Show complete solution . Will rate your answer.
What is the pH at the equivalence point if 15.84 ml of ammonia (NH3) is titrated with 53.24 ml of 0.1966 M HCl? Assume the volumes are additive.
Calculate the pH at the halfway point and at the equivalence point for each of the following titrations a. 100.0 ml of 0.70M HC7H5O2 (Ka= 6.4x10^-5) titrated by 0.10 M NaOH pH at the halfway point = ______? pH at the equivalence point = _____? b. 100.0ml of 0.70M C2H5NH2 (Kb= 5.6x10^-4) titrated by 0.60M HN03 pH at the halfway point = ______? pH at the equivalence point = _____? c. 100.0 ml of 0.70M HCL titrated by 0.15m NaOH...
4) Calculate the pH at the equivalence point for the titration below: 150 mL 0.10 M HCI against 75 mL of 0.20 M NH3 4) Calculate the pH at the equivalence point for the titration below: 150 mL 0.10 M HCI against 75 mL of 0.20 M NH3
4. Calculate the percent ionization and pH of the following solutions: a. 0.175 M NH3 Kg = 1.8 x 10-5. b. 35.0 mL of 0.175 M NH3 is mixed with 15.0 mL of 0.10 M HCl. c. 50.0 mL of 0.275 M acetic acid is mixed with 100. mL of 0.100 M HCl. d. The equivalence point of a 50.0 mL solution of 0.120 M acetic acid titrated by 0.100 M NaOH.
Sketch a pH titration curve if 100.0 mL of 0.125 M NH3 solution is titrated with 0.15M HCl. Note the following three items on the curve. K, for NHa 1.8 x 105 pH= Calculate the starting pH (no HCl added) vol (mL)= Calculate the volume of HCl added to reach the equivalence point pH = Calculate the pH at the equivalence point.
4. Please calculate the resulting pH when 25 mL of 0.10 M acetic acid is titrated with 10 mL of 0.10 M NaOH (K. acetic acid = 1.8 x 10-5): CH,COOH (aq) + NaOH(aq) → CH3COONa (aq) + H20 (1)
25 mL of 0.080 M solution of ammonia was titrated with 0.10 M HCl solution until the pH attained "saturation value". Draw the titration curve illustrating dependence between pH and volume of the HCI solution. Give the required answers in the boxes provided. Ko of NH3 = 1.8 x 10-5 Give initial pH value (must be correct within 0.5 pH unit) .... pH Circle whether the pH value at the equivalence point is lower than 7, equal to 7 or...