J.R. S. answered • 04/18/22
Ph.D. University Professor with 10+ years Tutoring Experience
KOH + HBrO ==> H2O + KBrO ... balanced equation
moles KOH = 10.0 ml x 1 L/1000 ml x 0.150 mol/L = 1.50x10-3 moles = 0.00150 mols
moles HBrO = 20.0 ml x 1 L/1000 ml x 0.300 mol/L = 6.00x10-3 moles = 0.00600 mols
KOH + HBrO ==> H2O + KBrO
0.0015.....0.00600..........0..............0...........Initial
-0.0015....-0.0015........................+0.0015..Change
0...............0.0045.........................0.0015....Equilibrium
Final volume = 10 ml + 20 ml = 30 mls = 0.030 L
Final [HBrO] = 0.0045 mols / 0.030 L = 0.15 M
Final [BrO-] = 0.0015 mols / 0.030 L = 0.05 M
Using the Henderson Hasselbalch equation:
pH = pKa + log [BrO-] / [HBrO]
pH = ?
pKa =-log 2.5x10-9 = 8.60
pH = 8.60 - log (0.05/0.15)
pH = 8.60 - 0.48
pH = 8.12
J.R. S.
04/11/24
Connor C.
Hello, I think I have found another discrepancy in your answers. The real answer to this question is 8.12, not 4.71. Let's instead use Henderson Hasselbach to answer this question. First, find the moles of each reactant (Which you did correctly state) KOH = .0015 mol HBrO = .00600 mol then, using Henderson Hasselbach: pH = -log(pKa) + log(base/acid) pH = -log(2.5*10^-9) + log(.00150/.006 -.0015) For anyone wondering why we use .006 - .0015 for our base in the logarithm, remember that in our ICE table (which he also does correct) that you are left with .0045 mol! This gets you an answer of 8.12. Stay sharp students!04/11/24