The pH of 0.50 M HBrO(aq) is 4.50. Calculate the change in pH when 6.50 g of sodium hypobromite is added to 110. mL of the solution. Ignore any change in volume. Ka of HBrO is 2E-9
I found the amount of moles used for each (.0547 mol NaBrO, and .055 mol HBrO soln)
I'm not totally sure how to find the new pH. Do I set up an ICE chart? If I do...I'm not sure how to set up the expression for the addition of NaBrO to HBrO...
I am trying to use pH=pKa + log(acid/base) to see where that gets me...I'm attempting to find the [base] using the given pH and pKa of 8.69 (for HBrO)...then use the concentration to find the new pH...
You can use the Henderson-Hasselbalch equation to calculate pH or you can set up the Ka expression and substitute (HBrO) where needed and use (NaBrO) as (BrO^-) in the Ka expression. Calculate H^+ and pH from that.
I am pretty stuck...not sure if I used the Henderson-Hasselbach the right way and I'm a little confused on how to set up the Ka expression...
Nevermind I got it...I just mixed up my variables a little. thanks!
First, a correction. The H-H equation is
pH = pKa + log (base/acid); your fraction is invered incorrectly.
Here are both ways:
HBrO ==> H^+ + BrO^-
Ka = 2 x 10^-9 = (H^+)(BrO^-)/(HBrO)
Plug in (BrO^-) 6.50grams/molar mass NaBrO and that divided by 0.110 L. (HBrO) = 0.5 M and solve for H^+ then convert to pH.
The HH equation is easier.
pKa = 8.70 rounded a little but you need to confirm that.
pH = pKa + log (base/acid)
Plug in Ka, (base) = 6.5/molar mass NaBrO and divide by 0.110 L. For acid plug in 0.5 M. I get pH about 7.7 or so.
To find the change in pH when sodium hypobromite (NaBrO) is added to the HBrO solution, you can indeed set up an ICE (initial, change, equilibrium) chart. Here's how you can do it:
1. Calculate the initial concentration of HBrO (aq):
- Given that the concentration of HBrO is 0.50 M and the volume is 110 mL, you can convert the volume to liters by dividing by 1000: 110 mL ÷ 1000 = 0.110 L.
- Since the concentration is given in moles per liter (M), the initial moles of HBrO can be calculated as: 0.50 M × 0.110 L = 0.055 mol.
2. Calculate the moles of NaBrO added:
- Given that the mass of sodium hypobromite is 6.50 g and its molar mass is 102.89 g/mol, you can calculate the moles of NaBrO as: 6.50 g ÷ 102.89 g/mol = 0.063 mol.
3. Calculate the change in moles of HBrO and NaBrO:
- Since NaBrO reacts with HBrO in a 1:1 ratio according to the balanced equation, the moles of HBrO will decrease by 0.063 mol, and the moles of NaBrO will increase by 0.063 mol.
4. Set up the expression for the addition of NaBrO to HBrO:
- Since HBrO is a weak acid, you can set up the equilibrium expression using the dissociation reaction: HBrO ⇌ H+ + BrO-
- The initial concentration of H+ is negligible, so you can omit it from the expression.
- The equilibrium concentration of HBrO can be calculated as: (0.055 - 0.063) mol ÷ 0.110 L = -0.008 M.
- The equilibrium concentration of BrO- after the reaction can be calculated as: (0.063 mol) ÷ 0.110 L = 0.573 M.
5. Calculate the new concentration of HBrO and BrO-:
- The initial concentration of HBrO (equilibrium concentration before adding NaBrO) is 0.055 M.
- The final concentration of HBrO (equilibrium concentration after adding NaBrO) is (0.055 - 0.063) M = -0.008 M. (Note: Since the concentration cannot be negative, we take its absolute value)
6. Calculate the new pH:
- To find the new pH, you need to calculate the concentration of H+ ions in the solution. Since HBrO is a weak acid, you can use the Ka value provided (2E-9) to calculate the concentration of H3O+ ions in the solution.
- The equation for the dissociation of HBrO is: HBrO ⇌ H+ + BrO-
- The Ka expression is: Ka = [H+][BrO-] / [HBrO]
- Substituting the known values, we have: 2E-9 = [H+](0.573) / (-0.008)
- Rearrange the equation to solve for [H+]: [H+] = (2E-9)(-0.008) / (0.573)
- Calculate the value of [H+].
7. Finally, calculate the new pH using the equation pH = -log[H+].
By following these steps, you should be able to calculate the change in pH when 6.50 g of sodium hypobromite is added to 110 mL of the HBrO solution.