Hello, Could you please provide a detail explanation as to how to find each of these...
Hello,
Could you please provide a detail explanation as to how to find each of these steps.
The ionization constant of a very weak acid, HA, is 5.2×10-9. Calculate the equilibrium concentrations of H3O+, A-, and HA in a 0.040 M solution of the acid.
Determine the concentrations of all species at equilibrium and the solution pH.
[H3O+] = mol/L
[A-] = mol/L
[HA] = mol/L
pH =
Homework Answers
![Ka for HA= 5.2 x 10 9 H3o+ НА + H2O оочом + 4- нро - Х (ооцо-х) - л + ka- [Hot] [RO] THAJ xxx = 512 X10-9 oo Чо- х? = зох)о-1](http://img.homeworklib.com/questions/18454270-7412-11ea-ae79-1ff53c73fcf1.png?x-oss-process=image/resize,w_560)
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Hello,
Could you please provide a detail explanation as to how to find
each of these...
The ionization constant of a very weak acid, HA, is 8.8x10-9. Calculate the equilibrium concentrations of...
The ionization constant of a very weak acid, HA, is 8.8x10-9. Calculate the equilibrium concentrations of H2O, A', and HA in a 0.040 M solution of the acid. Determine the concentrations of all species at equilibrium and the solution pH. [H30]= mol/L [A] = moll mol/L [HA] = pH =
How do you calculate the pH of a solution containing a weak acid or a weak...
How do you calculate the pH of a solution containing a weak acid or a weak base? Please write down the general formula for finding the H3O+ ion concentrations in case of a weak acid Please explain the common-ion effect in terms of Le Châtelier. You might want to use the general chemical equation for an acid ionization HA + H2O ↔ A- + H3O+ to help you explain the phenomenon. How do you calculate the degree of ionization? Please...
please help me please i dont have questions remaining anymore A solution is prepared by dissolving 0.060 mol of iodoacet...
please help me please i dont
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A solution is prepared by dissolving 0.060 mol of iodoacetic acid (Kg = 6.68x10-4) and 0.040 mol of sodium iodoacetate in water to a total solution volume of 1.00 L. First, calculate a pH for the solution by assuming that the concentrations of ICH CO2H and ICH CO2 are equal to the formal concentrations of the substances dissolved. Submit Answer Tries 0/5 Next, calculate the pH for the solution based...
Question 1 HA is a weak acid. Its ionization constant, Ky, is 5.1 x 10-13. Calculate...
Question 1 HA is a weak acid. Its ionization constant, Ky, is 5.1 x 10-13. Calculate the pH of an aqueous solution with an initial NaA concentration of 0.053 M. Question 2 We place 0.134 mol of a weak acid, HA, in enough water to produce 1.00 L of solution. The final pH of the solution is 1.18. Calculate the ionization contant, K. of HA. Question 3 We place 0.577 mol of a weak acid, HA, and 13.6 g of...
Question 1 : HA is a weak acid. Its ionization constant, Ka, is 1.2 x 10-13....
Question 1 : HA is a weak acid. Its ionization constant, Ka, is 1.2 x 10-13. Calculate the pH of an aqueous solution with an initial NaA concentration of 0.075 M. Question 2 : We place 0.143 mol of a weak acid, HA, in enough water to produce 1.00 L of solution. The final pH of the solution is 1.28 . Calculate the ionization contant, Ka, of HA. Question 3 : We place 0.661 mol of a weak acid, HA,...
- Given this concentration n ratio what is the pH (show work), Ka and percent ionization...
- Given this concentration n ratio what is the pH (show work),
Ka and percent ionization of this species
( HA+H2O← → A- + H3O+ , acid. Initial concentration: 0.01
mol/L)
HA: 9.77x10^-3 mol/L
H2O: 55.6 mol/L
A- : 2.34x10^-4 mol/L
H3O+: 2.34x10^-4 mol/L
- Also if we increase the initial concentration how does this
affect the ph n the Ka such as increasing it to .1mol/L
- is you have initial concentration of .001 if u increase
strength what...
We place 0.612 mol of a weak acid, HA, and 12.9 g of NaOH in enough...
We place 0.612 mol of a weak acid, HA, and 12.9 g of NaOH in enough water to produce 1.00 L of solution. The final pH of this solution is 4.42 . Calculate the ionization constant, Ka, of HA.
We place 0.577 mol of a weak acid, HA, and 13.6 g of NaOH in enough...
We place 0.577 mol of a weak acid, HA, and 13.6 g of NaOH in enough water to produce 1.00 L of solution. The final pH of this solution is 4.50 . Calculate the ionization constant, Ka, of HA.
We place 0.565 mol of a weak acid, HA, and 11.6 g of NaOH in enough...
We place 0.565 mol of a weak acid, HA, and 11.6 g of NaOH in enough water to produce 1.00 L of solution. The final pH of this solution is 4.88 . Calculate the ionization constant, Ka, of HA.
We place 0.516 mol of a weak acid, HA, and 12.0 g of NaOH in enough...
We place 0.516 mol of a weak acid, HA, and 12.0 g of NaOH in enough water to produce 1.00 L of solution. The final pH of this solution is 4.20. Calculate the ionization constant, Ka, of HA.
The ionization constant of a very weak acid, HA, is 8.8x10-9. Calculate the equilibrium concentrations of H2O, A', and HA in a 0.040 M solution of the acid. Determine the concentrations of all species at equilibrium and the solution pH. [H30]= mol/L [A] = moll mol/L [HA] = pH =
please help me please i dont
have questions remaining anymore
A solution is prepared by dissolving 0.060 mol of iodoacetic acid (Kg = 6.68x10-4) and 0.040 mol of sodium iodoacetate in water to a total solution volume of 1.00 L. First, calculate a pH for the solution by assuming that the concentrations of ICH CO2H and ICH CO2 are equal to the formal concentrations of the substances dissolved. Submit Answer Tries 0/5 Next, calculate the pH for the solution based...
Question 1 HA is a weak acid. Its ionization constant, Ky, is 5.1 x 10-13. Calculate the pH of an aqueous solution with an initial NaA concentration of 0.053 M. Question 2 We place 0.134 mol of a weak acid, HA, in enough water to produce 1.00 L of solution. The final pH of the solution is 1.18. Calculate the ionization contant, K. of HA. Question 3 We place 0.577 mol of a weak acid, HA, and 13.6 g of...
- Given this concentration n ratio what is the pH (show work),
Ka and percent ionization of this species
( HA+H2O← → A- + H3O+ , acid. Initial concentration: 0.01
mol/L)
HA: 9.77x10^-3 mol/L
H2O: 55.6 mol/L
A- : 2.34x10^-4 mol/L
H3O+: 2.34x10^-4 mol/L
- Also if we increase the initial concentration how does this
affect the ph n the Ka such as increasing it to .1mol/L
- is you have initial concentration of .001 if u increase
strength what...
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