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Cerrena [4.2K]
2 years ago
11

What volume of a 0.3300M solution of sodium hydroxide would br ruired to titrate 15.00 mL of 0.1500 M oalic Acid?

Chemistry
1 answer:
Ganezh [65]2 years ago
4 0

Answer:

3.41 mL

Explanation:

At equivalence point  from the reaction given,

Moles of Oxalic\ Acid = 2 × Moles of NaOH

Considering

Molarity_{Oxalic\ Acid}\times Volume_{Oxalic\ Acid}=2\times Molarity_{NaOH}\times Volume_{NaOH}

Given  that:

Molarity_{NaOH}=0.3300\ M

Volume_{NaOH}=?

Volume_{Oxalic\ Acid}=15.00\ mL

Molarity_{Oxalic\ Acid}=0.1500\ M

So,  

Molarity_{Oxalic\ Acid}\times Volume_{Oxalic\ Acid}=2\times Molarity_{NaOH}\times Volume_{NaOH}

2\times 0.3300\times Volume_{NaOH}=0.1500\times 15.00

Volume_{NaOH}=3.41\ mL

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Explanation:

Our solution if made of a solute named Tris

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This is the concentration of solution:

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Notice that mM = M . 1000  We convert from mM to M

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M = molarity (moles of solute in 1 L of solution, or mmoles of solute in 1 mL of solution). Let's determine the mmoles of Tris

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