How many milliliters of 0.100 m naoh are needed to completely neutralize 25.0 ml of 0.250 m h3po4 ?

1 answer:

6 0

C = n/V
n = C×V
n = 0,25M × 0,025L
n = 0,00625 mol H₃PO₄

1 mol................. 3 mol
H₃PO₄ ....+ ... 3NaOH ------> Na₃PO₄ + 3H₂O
0,00625............. X

X = (0,00625×3)/1 = 0,01875 mole of NaOH

C = n/V
V = n/C
V = 0,01875/0,1M
V = 0,1875 L = 187,5 mL NaOH

:)

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