Posted by reena on Monday, February 16, 2009 at 6:30pm.
First you must recognize that HI and KOH are strong acid + strong base respectively. The product is KI (a neutral salt) + H2O; therefore, the pH will depend entirely upon the reagent (HI or KOH) that is in excess. How do you determine the one in excess?
Calculate moles HI initially. That is M x L = ?
Calculate moles KOH initially. That is M x L = ?
Now subtract one from the other to see which one is in excess (obviously, the other one goes to zero since all of it will be used). Then moles/volume = M. Finally, if KOH is in excess, the pOH is the -log(OH^-) remaining or if HI is in excess, pH = -log(H^+) remaining.
Post your work if you get stuck.
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