Question

An aqueous solution of sodium hydroxide is standardized by titration with a 0.113 M solution of hydrolodic acid. If 15.7 mL of base are required to neutralize 14.4 mL of the acid, what is the molarity of the sodium hydroxide solution? M sodium hydroxide

Answer 1

Question 1

We can use the following formula

Va x Na = Vb x Nb

Va = 14.4 ml Na = 0.113 M

Vb = 15.7 ml Nb = ?

Substitute the values and calculate Nb

Nb = 14.4 x 0.113 / 15.7 = 0.10364 M

Hence concentration of the NaOH is 0.10364 M

Question 2

Volume of HI = 25.9 ml

Concentration of HI = 0.191 M

Moles of HI reacted = 25.9 x 0.191 / 1000 =  0.0049469‬ Moles

Moles of Ba(OH)2 required =   0.0049469‬ /2 =  0.00247345‬ Moles

Because Ba(OH)2 has 2 OH groups while HI has only one H+

Concentration of Ba(OH)2 =   0.00247345 x 1000 / 19.1 = 0.1295 M

Hence Concentration of Ba(OH)2 = 0.1295 M