Question

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?

Answer 1

Answer:

5.53 grams is the mass of NaOH in the mixture.

Explanation:

Let the mass of NaOH be x and $Ba(OH)_2$be y.

x + y = 10.1 g..[1]

Moles of NaOH = $\frac{x}{40 g/mol}$

Moles of $Ba(OH)_2=\frac{y}{171 g/mol}$

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

$n=1.53M\times 0.1078 L=0.164934 mol$

0.38709 moles neutralizes all the NaOH and $Ba(OH)_2$ present in solution.So, This manes that;

$\frac{x}{40 g/mol}+\frac{y}{171 g/mol}=0.164934 mol$

$171 x+40y=1128.15$ ..[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.