Asked by Matthew
25.0 mL of a 0.20 mol/L solution of potassium hydroxide is mixed with 25.0 mL of a 0.15 mol/L solution of hydrobromic acid. What is the pH of the resulting solution?
Answers
Answered by
DrBob222
KOH + HBr ==> KBr + H2O
millimols KOH = mL x M = 25.0 x 0.20 = 5.00
millimols HBr = 25.0 x 0.15 = 3.75
millimols KOH left over = 5.0 - 3.75 = 1.25
(KOH) remaining = millimoles/mL = 1.25/50.0 = ?
pOH = - log (KOH)
Then pOH + pH = pKw = 14. You know pKw and pOH, solve for pH.
Post your work if you get stuck.
millimols KOH = mL x M = 25.0 x 0.20 = 5.00
millimols HBr = 25.0 x 0.15 = 3.75
millimols KOH left over = 5.0 - 3.75 = 1.25
(KOH) remaining = millimoles/mL = 1.25/50.0 = ?
pOH = - log (KOH)
Then pOH + pH = pKw = 14. You know pKw and pOH, solve for pH.
Post your work if you get stuck.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.