Assume that you dissolve 10.1 g of a mixture of NaOH and Ba(OH)2 in 253.0 mL of water and titrate with 1.53 M hydrochloric acid. The titration is complete after 107.8 mL of the acid has been added. What is the mass (in grams) of NaOH in the mixture?

5.53 grams is the mass of NaOH in the mixture.

Explanation:

Let the mass of NaOH be x and be y.

x + y = 10.1 g..[1]

Moles of NaOH =

Moles of

Moles of HCl = n

Volume of HCl solution = 107.8 mL = 0.1078 L( 1 mL = 0.001 L)

Molarity of HCl solution = 1.53 M

0.38709 moles neutralizes all the NaOH and present in solution.So, This manes that;

..[2]

On solving [1] and [2] we get ;

x = 5.53 g

y = 4.57 g

5.53 grams is the mass of NaOH in the mixture.