Asked 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
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
Answers
Answered by
DrBob222
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.
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.
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