pH = 12.33
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
The pH of the resultant solution is 12.3.
The term pH refers to the acidity or alkalinity of the solution. From the question, we have the following information;
Number of moles of LiOH = 0.20 M * 35/1000 = 0.007 moles
Number of moles of HCl = 0.25 M * 23/1000 = 0.00575 moles
The equation of the reaction is; LiOH(aq) + HCl(aq) ----> LiCl(aq) + H2O(l)
We can see that HCl is the limiting reactant here hence LiOH is in excess by 0.00125 moles.
Concentration of excess LiOH = 0.00125 moles/(35.0 + 23.0 ml) * 10^-3
= 0.02 M
Now;
pOH = -log(0.02 M) = 1.7
pH = 14 - 1.7 = 12.3
Learn more about pH: https://brainly.com/question/4144781
