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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.

Respuesta :

aiirrx
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: 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,

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]M_1[/tex] = molarity of NaOH solution = 0.135 M

[tex]V_1[/tex] = volume of NaOH solution = 42.24 ml

[tex]M_2[/tex] = molarity of [tex]H_3PO_4[/tex] solution = ?M

[tex]V_2[/tex] = volume of [tex]H_3PO_4[/tex] solution = 25 ml

[tex]n_1[/tex] = valency of [tex]NaOH[/tex] = 1

[tex]n_2[/tex] = valency of [tex]H_3PO_4[/tex] = 3

[tex]1\times (0.135M)\times 42.24=3\times M_2\times 25[/tex]

[tex]M_2=0.076M[/tex]

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.

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