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

$1\times (0.135M)\times 42.24=3\times M_2\times 25$

$M_2=0.076M$

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.