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 clear explanation please

Answer 1

The equation

NaOH + HCl = H2O + NaCl

Ba(OH)2 + 2HCl = 2H2O + BaCl2

then... we need total OH-

total mmol of HCl = MV = 1.55*108.5 = 168.175 mmol of H+

then

mmol of OH- = 168.175 mmol

of which 1 mmol is NAOH and 2 mmol are from BA(OH)2

then

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

mol of NaOH + 2*mol of BA(OH)2 = mol of OH-

get MW

MW of NaOH = 40

MW of Ba(OH)2 = 171.3

now..

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

mol of NaOH + 2*mol of BA(OH)2 = 168.175 *10^-3

assume "x" is the mass of NAOH, and "y" the mass of Ba(OH)2

x+y = 10

mol of NaOH = mass/MW = x/40; mol of Ba(OH)2= mass/MW = y/171.3

x+ y = 10 g

x/40 + 2*y/171.3 = 168.175 *10^-3

y = 10-x substitte:

x/40 + 2*(10-x)/171.3  = 168.175 *10^-3

1/40*x + 2/171.3 *10 -  2/171.3x = 168.175 *10^-3

x(1/40 - 2/171.3) = 168.175 *10^-3-2/171.3 *10

x = (0.05142) / (1/40 - 2/171.3) = 3.8590

now

y = 10-3.8590 = 6.141

NaOH = 3.8590 g

Ba(OH)2 = 6.141 g