Water ionizes to a very small degree into hydronium ions, H3O+ , and hydroxide i

Question

Water ionizes to a very small degree into hydronium ions, H3O+ , and hydroxide ions , OH-: 2H2O(l) reverse arrows H3O+(aq) + OH-(aq) Kw = (H3O+) (OH-) = 1.00x10 -14 The very small value of the equilibrium constant , Kw, should give you an appreciation for how few water molecules actually ionize in pure water. . In neutral solutions, (H3O+) = (OH-) . In basin solutions, (H3O+) < (OH-) . In acidic solutions , ( H3O+) > ( OH -) But in all aqueous solutions, the product of the hydronium and hydroxide concentrations is equal to Kw. This, Kw allows you to calculate (H3O+) from (OH-) , or visa versa.
Part A- 5.00 x 10 -3 million of HBr are dissolved in water to make 14.0 L of solution. What is the concentration of hydroxide ions, (OH-) , in this solution? (OH-) =________value, units
Part B- 6.00 g of NaOH are dissolved in cold water to make 5.00L of solution. What is the concentration of hydronium ions, (H3O+) , in this solution? (H3O+) =________value, units Water ionizes to a very small degree into hydronium ions, H3O+ , and hydroxide ions , OH-: 2H2O(l) reverse arrows H3O+(aq) + OH-(aq) Kw = (H3O+) (OH-) = 1.00x10 -14 The very small value of the equilibrium constant , Kw, should give you an appreciation for how few water molecules actually ionize in pure water. . In neutral solutions, (H3O+) = (OH-) . In basin solutions, (H3O+) < (OH-) . In acidic solutions , ( H3O+) > ( OH -) But in all aqueous solutions, the product of the hydronium and hydroxide concentrations is equal to Kw. This, Kw allows you to calculate (H3O+) from (OH-) , or visa versa.
Part A- 5.00 x 10 -3 million of HBr are dissolved in water to make 14.0 L of solution. What is the concentration of hydroxide ions, (OH-) , in this solution? (OH-) =________value, units
Part B- 6.00 g of NaOH are dissolved in cold water to make 5.00L of solution. What is the concentration of hydronium ions, (H3O+) , in this solution? (H3O+) =________value, units The very small value of the equilibrium constant , Kw, should give you an appreciation for how few water molecules actually ionize in pure water. . In neutral solutions, (H3O+) = (OH-) . In basin solutions, (H3O+) < (OH-) . In acidic solutions , ( H3O+) > ( OH -) But in all aqueous solutions, the product of the hydronium and hydroxide concentrations is equal to Kw. This, Kw allows you to calculate (H3O+) from (OH-) , or visa versa.
Part A- 5.00 x 10 -3 million of HBr are dissolved in water to make 14.0 L of solution. What is the concentration of hydroxide ions, (OH-) , in this solution? (OH-) =________value, units
Part B- 6.00 g of NaOH are dissolved in cold water to make 5.00L of solution. What is the concentration of hydronium ions, (H3O+) , in this solution? (H3O+) =________value, units

Explanation / Answer

part A)

moles of HBr = 5.00 x 10^-3

volume = 14.0 L

molarity of H3O+   = moles / volume

                           = 5.00 x 10^-3 / 14

                          = 3.57 x 10^-4 M

[H3O+][OH-]=1.0 x 10^-14

3.57 x 10^-4 x [OH-]    = 1.0 x 10^-14

[OH-]   = 2.80 x 10^-11 M

part B)

moles of NaOH = mass / molar mass

                         = 6 / 40

                       = 0.15

molarity = moles / volume

             = 0.15 / 5

             = 0.03 M

[H3O+]   = 10^-14 / 0.,03

[H3O+] = 3.33 x 10^-13 M

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