How many mL of .150 M sodium hydroxide is required to neutralize 25 mL of .125 M of phosphoric acid?
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To find out the number of milliliters (mL) of a solution required to neutralize another solution, we need to use stoichiometry. The balanced chemical equation for the reaction between sodium hydroxide (NaOH) and phosphoric acid (H3PO4) is:
3 NaOH + H3PO4 -> Na3PO4 + 3 H2O
From the equation, we can see that for each molecule of phosphoric acid (H3PO4), we need 3 molecules of sodium hydroxide (NaOH) to completely neutralize it.
Given:
Volume of phosphoric acid solution (H3PO4) = 25 mL
Molarity of phosphoric acid solution (H3PO4) = 0.125 M
Molarity of sodium hydroxide solution (NaOH) = 0.150 M
To solve this problem, we can use the equation:
Molarity (M) = moles/Liters
First, let's calculate the moles of phosphoric acid (H3PO4):
Moles of H3PO4 = Volume (L) × Molarity (M)
= (25 mL ÷ 1000) L × 0.125 M
= 0.003125 moles
Since the stoichiometry of the reaction between H3PO4 and NaOH is 1:3, we know that we need three times as many moles of NaOH as moles of H3PO4:
Moles of NaOH = 3 × Moles of H3PO4
= 3 × 0.003125 moles
= 0.009375 moles
Now, let's find the volume of sodium hydroxide (NaOH) solution needed in mL:
Volume of NaOH solution (mL) = (Moles of NaOH ÷ Molarity of NaOH) × 1000
= (0.009375 moles ÷ 0.150 M) × 1000
= 62.5 mL
Therefore, you would need 62.5 mL of 0.150 M sodium hydroxide solution to neutralize 25 mL of 0.125 M phosphoric acid solution.