A 0.10 M phosphoric acid (H3PO4) solution is titrated with 0.15 M sodium hydroxide (NaOH). Calculate the volume of NaOH solution required to neutralize 50.0 mL of H3PO4 solution.
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To solve this problem, we need to use the equation for neutralization reactions:
acid + base → salt + water
In this case, phosphoric acid is the acid and sodium hydroxide is the base. The balanced equation for this reaction is:
H3PO4 + 3NaOH → Na3PO4 + 3H2O
From the equation, we can see that 1 mole of phosphoric acid reacts with 3 moles of sodium hydroxide, so the stoichiometry is 1:3.
First, we need to calculate the number of moles of phosphoric acid in 50.0 mL of the solution. We can do this using the given concentration and volume:
moles of H3PO4 = concentration × volume
= 0.10 M × 50.0 mL
= 0.010 mol
Since the stoichiometry is 1:3, we know that 1 mole of phosphoric acid reacts with 3 moles of sodium hydroxide. Therefore, the number of moles of sodium hydroxide required to neutralize the acid is:
moles of NaOH = 3 × moles of H3PO4
= 3 × 0.010 mol
= 0.030 mol
Finally, we can calculate the volume of the 0.15 M sodium hydroxide solution needed:
volume of NaOH solution = moles of NaOH / concentration
= 0.030 mol / 0.15 M
= 0.20 L
= 200 mL
Therefore, the volume of 0.15 M sodium hydroxide solution required to neutralize 50.0 mL of 0.10 M phosphoric acid solution is 200 mL.
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