A 35.0-ml sample of 0.20 m lioh is titrated with 0.25 m hcl. What is the ph of the solution after 23.0 ml of hcl have been added to the base?

Answer;

pH = 12.33

Explanation;

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