Answer to Question #119973 in General Chemistry for Cass

Question

3. Now you would like to make the same 1.0L solution with a pH of 11.5 using LiOH (strong), how many grams of Lithium Hydroxide are required?

Answer 1

pOH = 14 - pH = 14 - 11.5 = 2.5

[OH^-] = 10^-pOH = 10^-2.5 = 0.00316 M

n(OH^-) = V*C = 1*0.00316 = 0.00316 mol

n(LiOH) = n(OH^-) = 0.00316 mol

m(LiOH) = n*M = 0.00316*(7+16+1) = 0.00316*24 = 0.07584 g