You\'re looking for Ka, the dissociation constant, so you measure this by adding the
substance to water. 1. Write the balanced equation HBrO + H2O = BrO- + H30+ 2. the pH is
4.95, so you know its acidic and there has to have hydronium as a product. if the pH was >7.0,
there would be OH- instead. the pH is a measure of H ions, and to get the concentration, you
simply do 10^(-pH). 10^(-4.95) = 1.12E-5 3. The equation for Ka is concentration of
products/concentration of reactants. always look at the coefficients infront of the molecules in
the reaction. in this case, it is [BrO-)][H3O+]/[HBrO] *liquids and solids are never included in
this, only aqueous solutions and gases are* 4. Your concentration of H3O was found to be
1.12E-5, and because the H30 and BrO are in a 1:1 ration in the equation, the concentration for
BrO is also 1.12E-5. The concentration for HBrO as a reactant is given, but the final
concentration is 0.063-concentration of hydronium...but since the concentration of the
hydronium is so small, it doesnt at all affect the final answer. 5. your equation for Ka=((1.12E-
5)(1.12E-5))/(0.063-1.12E-5). The final answer for your Ka is 2.0E-9 Hope this helps, please
rate.
Solution
You\'re looking for Ka, the dissociation constant, so you measure this by adding the
substance to water. 1. Write the balanced equation HBrO + H2O = BrO- + H30+ 2. the pH is
4.95, so you know its acidic and there has to have hydronium as a product. if the pH was >7.0,
there would be OH- instead. the pH is a measure of H ions, and to get the concentration, you
simply do 10^(-pH). 10^(-4.95) = 1.12E-5 3. The equation for Ka is concentration of
products/concentration of reactants. always look at the coefficients infront of the molecules in
the reaction. in this case, it is [BrO-)][H3O+]/[HBrO] *liquids and solids are never included in
this, only aqueous solutions and gases are* 4. Your concentration of H3O was found to be
1.12E-5, and because the H30 and BrO are in a 1:1 ration in the equation, the concentration for
BrO is also 1.12E-5. The concentration for HBrO as a reactant is given, but the final
concentration is 0.063-concentration of hydronium...but since the concentration of the
hydronium is so small, it doesnt at all affect the final answer. 5. your equation for Ka=((1.12E-
5)(1.12E-5))/(0.063-1.12E-5). The final answer for your Ka is 2.0E-9 Hope this helps, please
rate..
Youre looking for Ka, the dissociation constant.pdf
1. You're looking for Ka, the dissociation constant, so you measure this by adding the
substance to water. 1. Write the balanced equation HBrO + H2O = BrO- + H30+ 2. the pH is
4.95, so you know its acidic and there has to have hydronium as a product. if the pH was >7.0,
there would be OH- instead. the pH is a measure of H ions, and to get the concentration, you
simply do 10^(-pH). 10^(-4.95) = 1.12E-5 3. The equation for Ka is concentration of
products/concentration of reactants. always look at the coefficients infront of the molecules in
the reaction. in this case, it is [BrO-)][H3O+]/[HBrO] *liquids and solids are never included in
this, only aqueous solutions and gases are* 4. Your concentration of H3O was found to be
1.12E-5, and because the H30 and BrO are in a 1:1 ration in the equation, the concentration for
BrO is also 1.12E-5. The concentration for HBrO as a reactant is given, but the final
concentration is 0.063-concentration of hydronium...but since the concentration of the
hydronium is so small, it doesnt at all affect the final answer. 5. your equation for Ka=((1.12E-
5)(1.12E-5))/(0.063-1.12E-5). The final answer for your Ka is 2.0E-9 Hope this helps, please
rate.
Solution
You're looking for Ka, the dissociation constant, so you measure this by adding the
substance to water. 1. Write the balanced equation HBrO + H2O = BrO- + H30+ 2. the pH is
4.95, so you know its acidic and there has to have hydronium as a product. if the pH was >7.0,
there would be OH- instead. the pH is a measure of H ions, and to get the concentration, you
simply do 10^(-pH). 10^(-4.95) = 1.12E-5 3. The equation for Ka is concentration of
products/concentration of reactants. always look at the coefficients infront of the molecules in
the reaction. in this case, it is [BrO-)][H3O+]/[HBrO] *liquids and solids are never included in
this, only aqueous solutions and gases are* 4. Your concentration of H3O was found to be
1.12E-5, and because the H30 and BrO are in a 1:1 ration in the equation, the concentration for
BrO is also 1.12E-5. The concentration for HBrO as a reactant is given, but the final
concentration is 0.063-concentration of hydronium...but since the concentration of the
hydronium is so small, it doesnt at all affect the final answer. 5. your equation for Ka=((1.12E-
5)(1.12E-5))/(0.063-1.12E-5). The final answer for your Ka is 2.0E-9 Hope this helps, please
rate.
![You're looking for Ka, the dissociation constant, so you measure this by adding the
substance to water. 1. Write the balanced equation HBrO + H2O = BrO- + H30+ 2. the pH is
4.95, so you know its acidic and there has to have hydronium as a product. if the pH was >7.0,
there would be OH- instead. the pH is a measure of H ions, and to get the concentration, you
simply do 10^(-pH). 10^(-4.95) = 1.12E-5 3. The equation for Ka is concentration of
products/concentration of reactants. always look at the coefficients infront of the molecules in
the reaction. in this case, it is [BrO-)][H3O+]/[HBrO] *liquids and solids are never included in
this, only aqueous solutions and gases are* 4. Your concentration of H3O was found to be
1.12E-5, and because the H30 and BrO are in a 1:1 ration in the equation, the concentration for
BrO is also 1.12E-5. The concentration for HBrO as a reactant is given, but the final
concentration is 0.063-concentration of hydronium...but since the concentration of the
hydronium is so small, it doesnt at all affect the final answer. 5. your equation for Ka=((1.12E-
5)(1.12E-5))/(0.063-1.12E-5). The final answer for your Ka is 2.0E-9 Hope this helps, please
rate.
Solution
You're looking for Ka, the dissociation constant, so you measure this by adding the
substance to water. 1. Write the balanced equation HBrO + H2O = BrO- + H30+ 2. the pH is
4.95, so you know its acidic and there has to have hydronium as a product. if the pH was >7.0,
there would be OH- instead. the pH is a measure of H ions, and to get the concentration, you
simply do 10^(-pH). 10^(-4.95) = 1.12E-5 3. The equation for Ka is concentration of
products/concentration of reactants. always look at the coefficients infront of the molecules in
the reaction. in this case, it is [BrO-)][H3O+]/[HBrO] *liquids and solids are never included in
this, only aqueous solutions and gases are* 4. Your concentration of H3O was found to be
1.12E-5, and because the H30 and BrO are in a 1:1 ration in the equation, the concentration for
BrO is also 1.12E-5. The concentration for HBrO as a reactant is given, but the final
concentration is 0.063-concentration of hydronium...but since the concentration of the
hydronium is so small, it doesnt at all affect the final answer. 5. your equation for Ka=((1.12E-
5)(1.12E-5))/(0.063-1.12E-5). The final answer for your Ka is 2.0E-9 Hope this helps, please
rate.](https://image.slidesharecdn.com/yourelookingforkathedissociationconstant-230629172147-cc10593d/85/Youre-looking-for-Ka-the-dissociation-constant-pdf-1-320.jpg)
