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

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,  

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

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

$Volume_{NaOH}=3.41\ mL$