Hi
If I had 40 mL of 0.1 M HOBr and mixed that with 10 mL of 0.5 M Ca(OH)2 how would I go about calclating the pH? (i don't know if titration is right because equivalence, halfway pts, etc.. are not specified)..
Thank you very much.
2HOBr + Ca(OH)2 ==> Ca(OBr)2 + 2H2O
Just an ordinary acid + base reaction to give a salt + water.
moles HOBr = M x L = 0.040 x 0.1 = 0.004 moles.
moles Ca(OH)2 = M x L = 0.01 x 0.5 = 0.005
Set up an ICE chart so you can see what is going on.
If 0.004 mole HOBr react, it will use 0.002 mole Ca(OH)2 and produce 0.002 mole of the salt. You will have an excess of Ca(OH)2 and if I didn't goof there will be 0.003 moles Ca(OH)2 remaining. The volume will be 50 mL. Calculate the OH^- from that and pOH.
Don't forget there are TWO OH ions per mole Ca(OH)2.
To calculate the pH of the solution resulting from mixing 40 mL of 0.1 M HOBr and 10 mL of 0.5 M Ca(OH)2, you need to consider the reaction between the two substances and the resulting ions in solution.
First, let's write the balanced equation for the reaction between HOBr and Ca(OH)2:
HOBr + Ca(OH)2 -> H2O + CaBr2
Since Ca(OH)2, a strong base, reacts with HOBr, a weak acid, it will undergo complete neutralization. As a result, all calcium hydroxide will dissociate into calcium and hydroxide ions:
Ca(OH)2 -> Ca2+ + 2OH-
At the same time, the weak acid HOBr will partially dissociate:
HOBr -> H+ + OBr-
Now, we can determine the initial concentration of hydroxide ions (OH-) and hypobromous acid (HOBr). To do this, we need to use the given volume and concentration of each solution.
For 10 mL of 0.5 M Ca(OH)2:
- Moles of OH- ions = volume (L) x concentration (mol/L) = 0.01 L x 0.5 mol/L = 0.005 mol
For 40 mL of 0.1 M HOBr:
- Moles of HOBr = volume (L) x concentration (mol/L) = 0.04 L x 0.1 mol/L = 0.004 mol
Since the balanced equation shows a 1:1 stoichiometric ratio between OH- ions and HOBr, we can see that OH- is in excess. This means that all of the HOBr will react, and there will be some hydroxide ions left over.
To calculate the remaining OH- ions, subtract the moles of HOBr reacted from the initial moles of OH- ions:
Remaining OH- ions = initial moles of OH- ions - moles of HOBr reacted = 0.005 mol - 0.004 mol = 0.001 mol
Now we can calculate the concentration of OH- ions in the final solution. Divide the remaining moles of OH- ions by the total volume of the solution (10 mL + 40 mL = 50 mL = 0.05 L):
OH- concentration = remaining moles of OH- ions / total volume (L) = 0.001 mol / 0.05 L = 0.02 M
The OH- concentration can be used to calculate the pOH with the equation:
pOH = -log(OH- concentration) = -log(0.02) ≈ 1.7
Since the solution resulted in an excess of OH- ions, we can consider it as a basic solution. To find the pH, we can use the equation:
pH = 14 - pOH = 14 - 1.7 ≈ 12.3
Therefore, the pH of the resulting solution is approximately 12.3.