<p>can you explain the following question?</p>
<p>Which of the following solutions is the product of the neutralization reaction between 10 ml 0.2 M KOH and 10 ml 0.2 M HI?</p>
<p>a) 0.1 M KI (answer)
b)0.2 M KI
c)0.4 KI</p>
<p>The reaction will be as follows:</p>
<p>KOH + HI --> H2O + KI</p>
<p>That part is fairly simple. Next, you want to find the molarity. To do that, you could first use stoichiometry in order to figure out how many moles of KI would be formed. Since, as you can see from the equation I wrote, for every 1 mole of KOH put in, 1 mole of KI will be formed. In this case, since KOH and HI are in equimolar concentrations, you don't need to worry about limiting reactant. So, once you've found out how many moles of KI are formed (the answer to this is .002), you will divide this by the total volume (.01 L + .01 L = .02 L total). So, you get that the answer is .002/.02, or .1M KI. Any questions?</p>