<h3><u>Answer;</u></h3>
pH = 12.33
<h3><u>Explanation;</u></h3>
The equation of reaction is :
LiOH(aq) + HCl(aq) --> LiCl(aq) + H2O(l)
Reactants left after the titrant is added;
Total Moles LiOH;
= 0.035L LiOH × (0.2moles/L)
= 0.007moles of LiOH
Moles of HCl;
= 0.023L HCl × (0.25moles/L)
= 0.00575moles HCl is the limiting reagent
Reacting amount of moles of LiOH;
= 0.0575 moles HCl *(1mole LiOH/1moles HCl)
=0.00575 moles LiOH (reacted)
Moles of LiOH left;
= 0.007moles total - 0.00575moles that react
= .00125 moles of LiOH (left)
LiOH is a strong base, which means that it ionizes completely.
0.00125moles LiOH *(moles/0.058L) = 0.02155M of LiOH
LiOH(aq) --> Li+(aq) + OH-(aq)
[LiOH] = [OH-] = 0.02155 M
pOH = -log[OH-]
pOH = -log(0.02155)
pOH= 1.67
pH = 14 - pOH
pH = 14 - 1.67
pH = 12.33