1 If 15 ml of 300 M HCl is used to titrate 224 ml of NaOH wh

1) If 15 ml of .300 M HCl is used to titrate 22.4 ml of NaOH, what is the molarity of the NaOH?

2) If 1.50 grams of a solid acid (monoprotic) is dissolved in water and then titrated with the base from question one, it takes 32.0 ml of the base fro the titration. What is the molar mass of the acid?

Solution

1). balancing the equivalence we get

n1v1=n1v2

thus

15*0.3=22.4*M

so molarity of NaOH=0.2M

2).

mole of acid= 32*0.2*10^-3

= 6.4*10^-3

thus 1.5 /Mol. wt= 6.4*10^-3

so mol. wt.

=234.375 gm/mol