Question

A standard solution of 86% (w/w) H3PO4 (specific density = 1.71 g/mL) is to be diluted to get a1200 mL H3PO4 solution at 4.00 M. How many mL of the standard H3PO4 solution are required?

Answer 1

Answer: 319 ml

Step-by-step explanation:

Given : 86 g of
`H_3PO_4` is dissolved in 100 g of solution.

Density of solution = 1.71 g/ml

Volume of solution=
`\frac{\text {mass of solution}}{\text {density of solution}}=(100g)/(1.71g/ml)=58.5ml`

moles of
`H_3PO_4=\frac{\text {given mass}}{\text {Molar mass}}=(86g)/(98g/mol)=0.88mol`

`\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution in L}}` .....(1)

Molarity of standard
`H_3PO_4` solution =
`(0.88* 1000)/(58.5ml)=15.04M`

To calculate the volume of acid, we use the equation given by neutralization reaction:

`M_1V_1=M_2V_2`

where,

`M_1\text{ and }V_1` are the molarity and volume of standard acid which is
`H_3PO_4`

`M_2\text{ and }V_2` are the molarity and volume of diluted acid which is
`H_3PO_4`

We are given:

`M_1=15.04M\\V_1=?mL\\\\M_2=4.00M\\V_2=1200mL`

Putting values in above equation, we get:

`15.04* V_1=4.00* 1200\\\\V_1=319mL`

Thus 319 ml of the standard
`H_3PO_4` solution are required.