A 35.0mL sample of .20M LiOH is titrated with .25M HCL.

Question

What is the pH of the solution after 23.0 mL of HCL have been added to the base?

(PLEASE SHOW WORK)

A. 1.26

B. 1.67

C. 12.33

D. 12.74

Answer 1

First write the reaction:

LiOH(aq) + HCl(aq) --> LiCl(aq) + H2O(l)

Use stoichiometry to find out how much reactant is left after the titrant is added.

.035L LiOH *(.2moles/L) = .007moles of LiOH total

.023L HCl *(.25moles/L) = .00575moles HCl is the limiting reagent

.0575 moles HCl *(1mole LiOH/1moles HCl) =.00575 moles LiOH react

.007moles total - .00575moles that react = .00125 moles of LiOH left

LiOH is a strong base, which means that it ionizes completly.

.00125moles LiOH *(moles/.058L) = .02155M of LiOH

LiOH(aq) --> Li+(aq) + OH-(aq)

[LiOH] = [OH-] = .02155M

pOH = -log[OH-]

pOH = -log(.02155)

pOH= 1.67

pH = 14 - pOH

pH = 14 - 1.67

pH = 12.33