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.
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.
2 answers
Don't forget there are TWO OH ions per mole Ca(OH)2.
1 answer