Asked by Acrinogy
A solution is prepared by mixing 4.62 grams of potassium hydroxide in 1.00 L of water. After that, 250.0 mL of a 0.100 M solution of HCl is added to the solution. What is the resulting pH of the final mixture?
Answers
Answered by
DrBob222
mols KOH = grams/molar mass = ?
mols HCl = M x L
Which is in excess. How much is the excess. If H in excess, (H^+) = mols/L solution. If OH in excess, ((OH^-) = mols/L. Then pH = -log(H^+) if H^+ is in excess. If OH is in excess it is pOH = -log(OH^-), the pH + pOH = pKw = 14. You will know pKw and pOH, solve for pH.
mols HCl = M x L
Which is in excess. How much is the excess. If H in excess, (H^+) = mols/L solution. If OH in excess, ((OH^-) = mols/L. Then pH = -log(H^+) if H^+ is in excess. If OH is in excess it is pOH = -log(OH^-), the pH + pOH = pKw = 14. You will know pKw and pOH, solve for pH.
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