128 mL of 1.50 x 10-2 M NaOH is used to titrate 225 mL of a monoprotic acid with unknown concentration to reach the equivalence point. What is the concentration of the original acid?

1.50 x 10-2 M

1.17 x 10-1 M

1.92 x 10-3 M

8.53 x 10-3 M

Question

128 mL of 1.50 x 10-2 M NaOH is used to titrate 225 mL of a monoprotic acid with unknown concentration to reach the equivalence point. What is the concentration of the original acid?
		
1.50 x 10-2 M
		
1.17 x 10-1 M
		
1.92 x 10-3 M
		
8.53 x 10-3 M

marty1162 · Answer

2 Al + 3 Cl2 → 2 AlCl3 

(58 g Al) / (26.98154 g Al/mol) x (2 mol AlCl3 / 2 mol Al) x (133.3411 g AlCl3/mol) = 
286.63 g AlCl3 in theory 

(203 g) / (286.63 g) = 0.708 = 71% yield