What is the equilibrium concentration of hydrogen ion in a solution of 0.15 M ammonia, NH3, Kb=1.8 × 10−5
[H+] = 6.10 × 10-12 M
What is the equilibrium concentration of hydrogen ion in a solution of 0.15 M ammonia, NH3,...
QUESTION 18 What is equilibrium concentration of NH4 in a solution of 0.45 M NH3 in water? Kb for NH3 - 1.8 * 10- 3.5 * 10-5M 4.2 * 10-3M 2.8 * 10-3M © 1.6* 10-5 M 8.1 x 10-6M
What is the concentration of ammonium ion in a 0.400 M solution of ammonia? The Kb value for ammonia is 1.80 x 10−5. 0.397 M 2.67 × 10-3 M 0.400 M 7.20 × 10-6 M 1.80 × 10-5 M
A 20.0-mL sample of 1.50 M NH3 is titrated with 0.15 M HCl. What is the pH of the solution after 12.00 mL of acid have been added to ammonia? Kb for NH3 = 1.8 × 10–5
A 20.0-mL sample of 1.50 M NH3 is titrated with 0.15 M HCl. What is the pH of the solution after 12.00 mL of acid have been added to ammonia? Kb for NH3 = 1.8 × 10–5 5.93 9.30 8.06 9.07 10.45
What is the [H+] of a solution containing 0.30 M NH3 and 0.15 M NH4NO3? Kb for NH3 = 1.8 times 10^-5
What are the equilibrium concentrations of NH3, NH4+, and OH- in a 0.95 M solution of ammonia? Kb = 1.8x10^-5 What is the pH of the solution? pH =
What is the pH of 6.0 M aqueous solution of NH3 (ammonia). It's Kb = 1.8 x 10-5.
Ammonia, NH3, is a weak base with a Kb value of 1.8×10^−5. Part A What is the pH of a 0.370 M ammonia solution? Part B What is the percent ionization of ammonia at this concentration?
Ammonia, NH3, is a weak base with a Kb value of 1.8×10−5. Part A What is the pH of a 8.50×10−2 M ammonia solution? Part B What is the percent ionization of ammonia at this concentration?
a) What is the pH of a solution that consist of 0.20 M ammonia, NH3, and 0.20 M ammonium choride, NH4Cl? (Kb for ammonia is 1.8 x 10-5) b) 1.25 g of benzoic acid (C6H5CO2H) and 1.25 g of sodium benzoate (NaC6H5CO2) are dissolved in enough water to make 250 mL solution. Calculate the pH of the solution using the Handerson-Hasselbach equation (Ka for benzoic acid is 6.3 x 10-5). c) What is the pH after adding 82 mg of...