Calculate the pH at the equivalence point for the titration of 1.0 M ethylamine, C2H5NH2, by 1.0 M perchloric acid, HClO4. (pKb for C2H5NH2 = 3.25)
Calculate the pH at the equivalence point for the titration of 1.0 M ethylamine, C2H5NH2, by...
1.Determine the pH during the titration of 36.6 mL of 0.304 M ethylamine (C2H5NH2 , Kb = 4.3×10-4) by 0.304 M HI at the following points. (a) Before the addition of any HI (b) After the addition of 16.1 mL of HI (c) At the titration midpoint (d) At the equivalence point (e) After adding 51.2 mL of HI b.Determine the pH during the titration of 61.4 mL of 0.450 M nitrous acid (Ka = 4.5×10-4) by 0.450 M NaOH...
6.) Determine the pH during the titration of 34.9 mL of 0.220 M ethylamine (C2H5NH2, Kb = 4.3×10-4) by 0.220 M HNO3 at the following points. (Assume the titration is done at 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 12.7 mL of HNO3 (c) At the titration midpoint (d) At the equivalence point (e) After adding 52.0 mL of HNO3
1. A 23.6 mL sample of 0.391 M ethylamine, C2H5NH2, is titrated with 0.315 M nitric acid. At the equivalence point, the pH is . Ethylamine Kb = 4.3X10-4 Nitrous acid Ka1 = 4.5x10-4 2. A 29.1 mL sample of 0.336 M ethylamine, C2H5NH2, is titrated with 0.276 M hydrochloric acid. After adding 52.4 mL of hydrochloric acid, the pH is . hydrochloric acid Ka1 = 3.5x10-8 Ethylamine Kb = 4.3X10-4
6. In the titration of 90.0 mL of 0.150 M ethylamine, C2H5NH2, with 0.100 M HCl, find the pH at each of the following point in the titration. C2H5NH2 has a Ko value of 6.4 x 104. After 60mL of HCl has been added
Determine the pH at the equivalence (stoichiometric) point in the titration of 33 mL of 0.22 M C2H5NH2(aq) with 0.16 M HCl(aq). The Kb of ethylamine is 6.5 x 10-4
Calculate the pH at the equivalence point for the titration of 0.251 M B: (a weak base with pkb = 5.27) with 0.251 M HCI.
Find the pH of the solution obtained when 32 mL of 0.087 M ethylamine, C2H5NH2, is titrated to the equivalence point with 0.15 M HCl. The value of Kb for ethylamine is 4.7 x 10-4. (in 3 s.f.)
A 22.8 mL sample of 0.242 M ethylamine, C2H5NH2, is titrated with 0.273 M hydrobromic acid. At the equivalence point, the pH is Use the Tables link in the References for any equilibrium constants that are required.
2. A 35.00-ml sample of a 0.0870 M solution of ethylamine, C2H5NH2 (Kb = 5.6 x 10), is titrated with 0.150 M HCI. (a) What is the equivalence volume of HCI? (b) Calculate the pH of the solution: (1) prior to the start of the titration. (ii) after addition of 10.15 mL of the 0.150 M HCI. (iii) after addition of 14.50 mL of the 0.150 M HCI. (iv) after addition of 6.00 mL of the HCl beyond the equivalence...
A) A 27.5 mL sample of 0.223 M ethylamine, C2H5NH2, is titrated with 0.301 M hydrobromic acid. After adding 29.5 mL of hydrobromic acid, the pH is . B) A 24.5 mL sample of 0.351 M diethylamine, (C2H5)2NH, is titrated with 0.302 M hydroiodic acid. At the titration midpoint, the pH is