Question

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

C2O4H2 + 2NaOh ? Na2C2O4 + 2H2O

Answer 1

3.41 mL

Step-by-step explanation:

At equivalence point from the reaction given,

Moles of
`Oxalic\ Acid` = 2 × Moles of NaOH

Considering

`Molarity_(Oxalic\ Acid)* Volume_(Oxalic\ Acid)=2* Molarity_(NaOH)* 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)* Volume_(Oxalic\ Acid)=2* Molarity_(NaOH)* Volume_(NaOH)`

`2* 0.3300* Volume_(NaOH)=0.1500* 15.00`

`Volume_(NaOH)=3.41\ mL`