I don't know how much you are to assume on this and how much you are to work out. Let's first assume what we need.
Since we don't know the (HOBr) at the beginning, we don't know if the final solution is before the equivalence point, at the eq. point, or after the eq. point. Let's assume we are before the eq. point. That means this is a buffered solution and we use the Henderson-Hasselbalch equation.
15 mL x 0.325M NaOH = 4.875 millimoles.
...........NaOH + HOBr ==> NaOBr + H2O
initial..4.875.....?.........0........0
change..-4.875....-4.875..+4.875..+4.875
equil......0......?-4.875..4.875..4.875
pH = pKa + log(base)/(acid)
7.48 = 8.70 + log[4.875/(acid)]
Solve for (acid). It is approximately 80 but you should do it more accurately.
The answer to the problem is you have 4.875 millimoles OBr^- and 80 mmoles HOBr(remember these are approximate) in the final solution and you should convert to moles because the problem asks for moles.
This leaves out the rest of the problem; i.e., how do we know this isn't the eq. point (calculated differently) or past the eq. point (calculated differently). If you don't need to worry about that we can drop it. If you are expected to know how to determine that, you will need to follow up with this question if you don't know how to do it.
15 ml of .325 M NaOH is delivered into a 35 ml HObr solution of unknown concentration.The pH of the final solution is measured to be 7.48.Ka for HOBr is 2.0 * 10^-9 at 25C.
Find the number of moles of OBr- and HOBr in the final solution
1 answer