Question

You dissolve 10.0 g of mixture of NaOH and Ba(OH)2 in 240.0 ml water and titrate with 1.55 M hydrochloric acid. The titration is complete after 108.5 ml of the acid has been acid. What is mass in grams of NaOH and Ba(OH)2

Answer 1

Work out moles of OH- ion that was present

H+ + OH- ------> H2O

moles H+ used = Molarity x L
moles H+ = 1.55 M x 0.1085 L
moles H+ = 0.168175 mol

Thus total moles OH- = 0.168175 mol

NaOH provides 1 OH-
let moles NaOH = a

Ba(OH)2 provides 2 OH-
let moles Ba(OH)2 = b

So equation 1

a + 2b = 0.168175 mol

Thus we get equation 2
a = 0.168175 - 2b

And we know

mass NaOH + mass Ba(OH)2 = 10 g

Mass = moles x molar mass

Mass NaOH = mol NaOH x 40.00 g/mol
                    = a(40.00)

Mass Ba(OH)2 = mol Ba(OH)2 x 171.34 g/mol
Mass Ba(OH) = b(171.34)


Equation 3
a(40) + b(171.34) = 10

substituting the value of a from equation 2 into equation 3 and solve for b

(0.168175 - 2b)(40) + 171.34b = 10

6.727 - 80b + 171.34b = 10

91.34b = 3.273

b = 3.273 / 91.34

b = 0.036

Then sub b into equation 2

a = 0.168175 - 2(0.036)

a = 0.168175 - 0.072

a = 0.096

a = moles NaOH = 0.096 mol
b = moles Ba(OH)₂ = 0.036 mol

mass NaOH = moles x molar mass
                   = 0.096 mol x 40.00 g/mol
                   = 3.84 g

mass Ba(OH)₂ = moles x molar mass
                        = 0.036 mol x 171.34 g/mol
                        = 6.16 g