posted by helen .
A solution of 0.178 M KOH(28.3 ml) is mixed with 28.9ml of 0.133 M HCl. Assuming that the final solution is the sum of the initial volumes, caculate:
a)the molarity of the K+ cation
b)the molatiry of the Cl- anion
c)the pH of the final solution
d)the pOH of the final solution
moles KOH = M x L = ?
moles HCl = M x L = ?
All K^+ come from KOH. All Cl^- come from HCl.
a. (K^+) = moles K^/L soln
b. (Cl^-) = moles Cl^-/L soln.
c. KOH + HCl ==> KCl + H2O
(H^+) or (OH^-) is determined by which reagent is in excess. Subtract the moles and see which is in excess and convert to pH or pOH.
Post your work if you get stuck.