35.0 mL of acid with an unknown concentration is titrated with 24.6 mL of 0.432 M base. What is the concentration of the acid? Assume the acid contributes 1 mole of (H+) ions/mole of acid and the base contributes 1 mole of (OH-) ions/mole of base.

3.29 M
0.615 M
0.304 M

Question

35.0 mL of acid with an unknown concentration is titrated with 24.6 mL of 0.432 M base. What is the concentration of the acid? Assume the acid contributes 1 mole of (H+) ions/mole of acid and the base contributes 1 mole of (OH-) ions/mole of base.
 
3.29 M
0.615 M
0.304 M

desmarie · Answer

@ganeshie8 
@sweetburger 
@.Sam.

desmarie · Answer

@Chevyrebel04 
@RhondaSommer

chevyrebel04 · Answer

its (b)

desmarie · Answer

m bad forgot to tell you that the correct answer is the last one but idk how they got that

chevyrebel04 · Answer

ohh sorry i got b for a answer but ok

rhondasommer · Answer

no direct answers please

aaronq · Answer

Because there is a 1:1 molar ratio (this is given in the question " the acid contributes 1 mole of (H+) ions/mole of acid and the base contributes 1 mole of (OH-) ions/mole of base."), you can use:

$\sf M_{acid}*V_{acid}=M_{base}*V_{base}$

where "M" represents molarity and "V", volume.

rearranging the formula to solve for molarity of base yields: 

$\sf M_{base}=\dfrac{M_{acid}*V_{acid}}{V_{base}}$