Chemistry - Stoichiometry - Titrations problem

can someone please help me on this problem? We did only one practice in class and the hw isn't similar to the only one we did and i cant solve it. Thank you soo much

What volume of .115 M HCLO4 solution is needed to neutralize 50.00 mL of .0875 M NaOH?

Ryan is wrong because he is using the dilutions equation and not using the stoichiometry concept correctly.

First write the balanced chemical equation.

HCLO4 + NaOH>>>>H2O + NaClO4
Recall that Molarity is moles/L
0.050L(0.0875 moles of NaOH/L)(1 mole of HClO4/1 mole of NaOH)(1 L/0.115 moles of HClO4)=0.03804L=38.04 mL

As you can see the stochiometry 1:1 , 1 mole of HClO4/1 mole of NaOH

Ryan has the same answer but the concept is wrong and will cost you future exercise. The dilution equation M1V1=M2V2 is meant to be use with the same chemical substance not with different species.

Tip: review the stochiometry concept.

This should be solvable using a Molarity * Volume equation since you are looking for neutralization, in which case the molarity of both the acid and base will equal each other and cancel out. The equation to be used is:

M1*V1 = M2*V2

If we let item 1 = HClO4
M1 = 0.115 M
V1 = ?

If we let item 2 = NaOH
M2 = 0.0875 M
V2 = 50 ml

This gives us
0.115M * X = 0.0875M * 50mL

Setting up to solve for X

X = (0.0875M * 50mL)/0.115M
solving M cancels out and X = 38.04mL

Hope this Helps!