Question

What is the concentration of a phosphoric acid solution of a 25.00 mL sample if the acid requires 42.24 mL of 0.135 M NaOH for neutralization?

Please explain your steps.

Answer 1

0.0760 m

do this by:

finding the moles of NaOH which will be 5.702 E -3 m

next find the moles of H3PO4 which will be 1.90 E -3 m
calulcate 25 ml sample molarity = 0.07603 m, just put 0.0760

Answer 2

Answer: The volume of 0.10 M NaOH required to neutralize 30 ml of 0.10 M HCl is, 30 ml.

Explanation:

According to the neutralization law,

`n_1M_1V_1=n_2M_2V_2`

where,

`M_1` = molarity of NaOH solution = 0.135 M

`V_1` = volume of NaOH solution = 42.24 ml

`M_2` = molarity of
`H_3PO_4` solution = ?M

`V_2` = volume of
`H_3PO_4` solution = 25 ml

`n_1` = valency of
`NaOH` = 1

`n_2` = valency of
`H_3PO_4` = 3

`1* (0.135M)* 42.24=3* M_2* 25`

`M_2=0.076M`

Therefore, the concentration of 0.076 M of phosphoric acid of a 25 ml is required to neutralize 42.24 ml of 0.135 M NaOH.