Hey guys! Can you please explain each step to the questions that you answer? I a
ID: 1051844 • Letter: H
Question
Hey guys! Can you please explain each step to the questions that you answer? I am having a little trouble with these questions and having a step by step explanation to the answers and how you got them would REALLY help me out. Also if possible, please write neatly :) I appreciate you guys so much. Thanks for all the hard work you guys put in.
24) Determine the pH of a 0.18 M H2CO3 solution. Carbonic acid is a diprotic acid whose Kal = 4.3 x 10-7 and Ka2= 5.6 x 10-11 25) Determine the pH of a solution that is 0.15 M HCIO2 (Ka-1.1 x 10-2) and 0.15 M HCIo (Ka-2.9 x 10-8) 26) Determine the concentration of CO32- ions in a 0.18 M H2CO3 solution. Carbonic acid is a diprotic acid whose Kal = 4.3 x 10-7 and Ka2 = 5.6 x 10-1 1 27. A solution containing AgNO3 is mixed with a solution of NaCl to form a solution that is 0.10 M in AgNO3 and 0.075 M in NaCl. What will happen once these solutions are mixed? Ksp (AgCl)= 1.77 x 10-10 A) Nothing will happen since the molar solubility of AgCl is higher than the solution concentrations. B) Silver chloride will precipitate out of solution, leaving an unsaturated solution of AgCI C) Silver chloride will precipitate out of solution, leaving a saturated AgCl solution D) Nothing will happen since NaCl and AgNO3 are both soluble compounds. E) There is not enough information to say anything about this solutionExplanation / Answer
Q27
A solution containing AgNO3 is mixed with a solution of NaCl to form a solution that is 0.10 M in AgNO3 and 0.075 M in NaCl. What will happen once these solutions are mixed? Ksp (AgCl) = 1.77 × 10-10.
Answer
(c) Silver chloride will precipitate out of solution, leaving a saturated AgCl solution.
Q24
Determine the ph of 0.18m H2C03 solution
H2CO3 + H2O <---> H3O+ + HCO3-
Initial: 0.18 M 0 0
Change: -x +x +x
Equilibrium: (0.18 - x) x x
Ka = [H3O+][HCO3-] / [H2CO3] = 4.3 x 10^-7
Ka1 = x2 / 0.18 = 4.3 x 10^-7
solve for x:
x = [(4.3 x 10-7)(0.18)]1/2 = 2.78 x 10-4 M = [H3O+]
pH = -log[H3O+]
Plug in the concentration
pH of H2CO3 solution = -log[2.78 x 10-4 M] = 3.56
q25
Determine the pH of a solution that is 0.15 M HClO2 (Ka=1.1E-2) and 0.15 M HClO (Ka=2.9E-8).
n equilibrium
HClO2 + H2O<<<>>>ClO2- + H3O+
H3O+=H+
Ka1 = [H+] [ClO2-] / [HClO2]
1.1 e-2 = [X] [X] / [0.15 -X]
1.1 e-2 [0.15 -X]= [X] [X]
using the quadratic equation
= 0.03549=[H+]
pH=-log[H+]= 1.449=1.45 -answer
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