15.0 mL of 0.50 M HCl is added to a 100.-mL sample of 0.462 M HNO2 (Ka for HNO2 = 4.0 x 10–4). What is the equilibrium concentration of NO2– ions?
We need at least 10 more requests to produce the answer.
0 / 10 have requested this problem solution
The more requests, the faster the answer.
15.0 mL of 0.50 M HCl is added to a 100.-mL sample of 0.462 M HNO2...
15.0 mL of 0.50 M NaOH is added to a 100.-mL sample of 0.442 M NH3 (Kb for NH3 = 1.8 x 10–5). What is the equilibrium concentration of NH4+ ions?
A 15.0 mL sample of 0.21 M nitrous acid (HNO2), Ka = 4.6x10-4) is titrated with 0.15 M NaOH. Calculate the pH after the addition of the following volumes of base: a. 0.00 mL b. 10.00 mL c. 21.00 mL d. 25.00 mL
For the following equation: The Ka value of HNO2 is 4.0 x 10-4. If the equilibrium concentration of HNO2 is 0.54 M. what is the equilibrium concentration of NO2/H30? 2 points HNO, + H, O NO + H20+ Your answer
Calculate the pH of the buffer that results from mixing 90.0 mL of 0.350 M HNO2 and 75.0 mL of 0.500 M NaNO2. (Ka = 4.60 x 10-4) a. Using the above initial buffer, calculate the new pH of the buffer when 15.0 mL of a 1.0 M solution of HCl is added. b. Using the above initial buffer, calculate the new pH of the buffer when 0.500 g of NaOH is added.
Calculate the pH after 4.0 mL of 0.50 M NaOH is added to 110.0 mL of a buffer made of 0.40 M CH3COOH and 0.50 M NaCH3COO. The Ka of acetic acid is 1.8 x 10.
Please show as much work as possible but also write neatly :) Thank you so much for your help! 3.3 x 10-1 M 15.0 mL of 0.50 M HCl is added to a 100.0 ml sample of 0.377 M HNO 2(Ko for HNO 2 - 4.0 x 104). What is the equilibrium concentration of NO 2-ions? 1.3 x 10-4 M 4.5 x 10-2M 2.0 x 10-3M none of these
What is the pH of a 0.50 M NO2- solution. The Ka of HNO2 is 4.5 x 10-4. a) 7.54 b) 8.52 c) 8.30 d) 7.76 e) 8.04
1.) If 15.0 mL of 3.60 M HCl (aq) are added to 10.0 mL of water, what is the concentration of the resulting solution? 2.) How many mL of water must be added to 25.0 mL of a 0.500 M NaCl solution to result in a solution that is 0.200 M NaCl? 3.) The net ionic equation for the reaction of aqueous solutions of sodium chloride and silver nitrate is...? 4.) How many mL of a 0.500 M sodium chloride...
References A 2.00 g sample of KCI is dissolved in 65.0 mL. of water. The resulting solution is then added to 15.0 mL. of a 0.430 M CaCl, (aq) solution. Assuming that the volumes are additive, calculate the concentrations of each ion present in the final solution. Concentration of K ions M Concentration of Ca ions Concentration of CI ions M Submit Answer Try Another Version 6 item attempts remaining
Question 1 (1 point) 12.0 mL of a 0.50 M Na2CO3 solution is added to a large test tube. Enough 0.50 M NaHCO3 solution is added to the test tube to give a final volume of 30.0 mL. What is the pH of the resulting buffer solution? H2CO3 has Ka1 = 4.3×10-7 and Ka2 = 5.6×10-11. Question 3 (1 point) A buffer solution is made by adding 15.0 mL of a 0.50 M Na2CO3 solution to 15.0 mL of a...