Question

Calculate the molarity of each aqueous solution:

a. 78.0 ml of 0.240 m naoh diluted to 0.250 l with water

b. 38.5 ml of 1.2 m hno3 diluted to 0.130 l with water

Answer 1

We know that when we dilute a solution the number of moles remains the same just the concentration changes [moles per unit volume]

So we can equate the moles before dilution and after dilution

moles = molarity X volume

a) Initial moles = final moles [NaOH]

$M1V1=M2V2$

M1= initial molarity = 0.240 M

V1= initial volume = 78 mL

M2= final molarity = ?

V2 = final volume = 0.250 L = 250mL

therefore

$M2=\frac{M1V1}{V2}=\frac{0.24X78}{250}=  0.075M$

final molarity = 0.075 M

b)

Initial moles = final moles [Nitric acid]

$M1V1=M2V2$

M1= initial molarity = 1.20 M

V1= initial volume = 38.5 mL

M2= final molarity = ?

V2 = final volume = 0.130 L = 130mL

therefore

$M2=\frac{M1V1}{V2}=\frac{0.1.2X738.5}{130}=  0.355M$

final molarity = 0.355 M